IN   MEMORIAM 
FLOR1AN  CAJOR1 


I 

' 


- 


MATHEMATICAL  TEXTS  FOR  SCHOOLS 

Edited  by  PERCEY  F.  SMITH,  Pn.D. 

Professor  of  Mathematics  in  the  Sheffield  Scientific  School 
of  Yale  University 


First  Course  in  Algebra 
Second  Course  in  Algebra 

Complete  School  Algebra 

By  H.  E.  HAWKES,  Pn.P.,  W.  A.  LIJBY,  A.B., 
and  F.  C.  TOUTON,  Pn.B. 

Plane  Geometry 

By  WILLIAM  BETZ,  M.  A.,  and  H.  E.WEBB,  A.B. 

Advanced  Algebra 

By  H.  E.  HAWKES,  PH.D. 

Plane  and  Spherical  Trigonometry  and  Four-Place 
Tables  of  Logarithms 

Plane  and  Spherical  Trigonometry 

Plane  Trigonometry  and  Four-Place  Tables  of 
Logarithms 

Four-Place  Tables  of  Logarithms 

By  W.  A.  GRANVILLE,  Pn.D. 


PLANE  GEOMETRY 


BY 


WILLIAM  BETZ,  A.M. 

VICE-PRINCIPAL  AND   HEAD  OF  THE   DEPARTMENT  OF  MATHEMATICS 
IN   THE  EAST   HIGH   SCHOOL,  ROCHESTER,  NEW  YORK 

AND 

HARRISON  E.  WEBB,  A.B. 

HEAD  OF   THE    DEPARTMENT   OF  MATHEMATICS   IN   THE   CENTRAL 

COMMERCIAL  AND  MANUAL   TRAINING   HIGH   SCHOOL 

NEWARK,  NEW  JERSEY 


WITH  THE  EDITORIAL  COOPERATION  OF 

PERCEY  F.  SMITH 

PROFESSOR  OF  MATHEMATICS  IN   THE   SHEFFIELD  SCIENTIFIC 
SCHOOL  OF   YALE    UNIVERSITY 


GINN  AND  COMPANY 

BOSTON  •  NEW  YORK  •  CHICAGO  •  LONDON 


COPYRIGHT,  1912,  BY  WILLIAM  BETZ,  HARRISON  E.WEBB 

AND  PERCEY  F.  SMITH 

ALL  RIGHTS  RESERVED 

912.6 


CAJOSI 


ZEfre 


GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PREFACE 

The  addition  of  another  text  to  the  long  list  of  existing 
geometry  texts  calls  for  an  explanation  of  the  motives  under- 
lying its  preparation. 

Briefly  stated,  this  text  aims  to  effect  a  compromise  between 
the  extreme  demands  of  certain  reformers  and  the  equally  un- 
tenable position  of  overconservative  writers.  Our  educational 
troubles  cannot  be  cured  by  a  complete  break  with  the  past,  nor 
by  ignoring  the  legitimate  demands  of  our  times.  Accordingly, 
the  authors  have  tried  to  endow  the  subject  with  life  and  reality, 
and  at  the  same  time  to  retain,  as  far  as  possible,  that  spirit  of 
careful  reasoning  for  which  geometry  has  always  been  famous. 

To  accomplish  this  double  purpose,  the  authors  depend 
mainly  on  the  following  features : 

1.  A  Preliminary  Course  precedes  the  Demonstrative  Course. 
It  had  long  been  the  experience  of  the  authors  that  many 
pupils  become  permanently  discouraged  at  the  very  beginning 
of  demonstrative  geometry  by  the  simultaneous  appearance 
of  too  many  difficulties.  The  Preliminary  Course  is  intended 
to  serve  as  a  preparation  for  formal  geometry  by  vitalizing  the 
content  of  all  definitions  by  abundant  illustration  and  discus- 
sion ;  by  cultivating  skill  in  the  use  of  ruler  and  compasses 
through  interesting  drawing  exercises  ;  by  presenting  exercises 
requiring  for  their  solution  simple  reasoning  and  inference ; 
and  by  developing  gradually  the  conviction  that  formal  proof 
is  necessary  for  further  advance. 

The  actual  time  required  by  the  Preliminary  Course  de- 
pends somewhat  on  the  class  and  on  the  mode  of  procedure. 
A  laboratory  plan  with  individual  notebooks  may  be  followed, 
or  the  exercises  may  be  solved  on  the  blackboard.  In  either  case 
the  Preliminary  Course  can  be  completed  in  five  or  six  weeks. 

911354 


vi  PKEFACE 

Careful  tests  in  many  classrooms  have  proved  that  the  intro- 
ductory work  results  in  an  actual  saving  of  time.  In  fact,  the 
entire  book  does  not  require  more  than  the  usual  allotted  time. 

2.  The  Demonstrative  Course  is  built  up  not  only  in  a  topical 
but  also  in  a  psychological  order.    The  authors  have  paid  great 
attention  to  a  natural,  progressive  order  of  topics,  and  have 
endeavored  to  arrange  a  sequence  of  theorems  which  should  be 
teachable.    The  assumptions  to  be  used  in  each  Book  are  sum- 
marized at  the  beginning,  and  each  difficult  topic  is  approached 
by  way  of  an  informal  discussion.    Consideration  of  the  incom- 
mensurable case  is  made  unnecessary  by  the  sequence  of  theo- 
rems and  by  assumptions  which  replace  the  usual  attempts  at 
demonstration.   Hypothetical  constructions  are  avoided.    More- 
over, the  whole  course  may  be  completed  without  a  formal  con- 
sideration of  the  theory  of  limits.    At  the  end  of  Book  V, 
however,  teachers  preferring  a  formal  treatment  of  limits  will 
find  a  clear  presentation  of  this  topic. 

3.  The  methods  embodied  in  the  text  aim  to  make  the  pupil 
independent  of  the  printed  page.    In  too  many  cases  students 
merely  verify  and  reproduce  the  statements  of  the  book.   A 
creative  spirit  is  absolutely  essential  in  geometry,  and  in  this 
text  the  attempt  is  made  to  cultivate  a  spirit  of  discovery  from 
the  very  beginning.    At  the  beginning  of  each  new  topic  com- 
plete or  nearly  complete  proofs  are  given.    As  the  student  ad- 
vances, proofs  are  either  omitted  (pp.  112,  148,  167),  or  are 
given  in  outline  (pp.  119,  273),  or  are  suggested  by  an  analy- 
sis (pp.  161,  268).    In  this  way,  even  after  a  rather  extensive 
discussion  in  the  classroom,  work  is  left  for  the  student  to  do. 

4.  The  various   types  of  exercises  receive  practicalbj   equal 
attention.    Constructions,  computations,  and  original  theorems 
will  be  found  throughout  the  text.    Many  of  these  carefully 
graded  exercises  can  be  worked  at  sight.    Special  attention  is 
called  to  the  use  of  composite  figures  (see  p.  116,  Ex.  8  ;  p.  117, 
Ex.9;  IV  141,  Ex.11;  p.  175,  Ex.14;  p.  213,  Ex.  11;  etc.). 
These  have  the  important  function  of  enabling  the  student  to 


PREFACE  vii 

make  his  own  discoveries.  To  aid  in  the  discovery  of  geo- 
metric relations,  tables  of  methods  are  given  from  time  to  time 
(see  pp.  76,  82,  85,  93,  118). 

5.  The  list  of  applied  problems  is  extensive  but  not  excessive. 
A  proper  balance  between  theory  and  practice  is  of  the  great- 
est importance,  if  the  demonstrative  work  is  not  to  be  seriously 
impaired.   Accordingly,  the  attempt  has  been  made  to  give  pref- 
erence to  such  applications  as  have  probably  come  within  the 
pupil's  range  of  experience.    Problems  requiring  an  unusual 
amount  of  explanation,  even  if  interesting  in  themselves,  have 
been  excluded. 

6.  The  text  provides  a  minor  and  a  major  course.    Thus  the 
teacher  may  omit,  without  modifying  the  remainder  of  the  work, 
any  or  all  of  the  following  pages :  Part  II  of  the  Preliminary 
Course  (pp.  53-68),  the  sections  on  coordinates  (pp.  177-180), 
the  trigonometric  work  (pp.  259-264),  and  the  formal  work  on 
limits  (pp.  309-316).    A  number  of  theorems  may  be  treated 
as  exercises,  for  example,  those  on  pp.  136-139,  152,  161,  225, 
268,  277-283. 

The  list  of  theorems  is  considerably  reduced,  but  it  is  entirely 
adequate  to  meet  the  requirements  of  higher  institutions. 

The  authors  note  with  pleasure  that  their  work  seems  to  be 
in  agreement  with  the  suggestions  of  various  associations  and 
committees,  notably  with  the  report  of  the  Committee  on  a 
National  Geometry  Syllabus. 

Acknowledgments  are  due  to  many  friends  for  valuable 
suggestions  and  criticisms.  In  particular,  the  authors  wish 
to  express  their  special  indebtedness  to  their  colleagues  in 
Rochester  and  Newark  for  their  interest  in  the  preparation  of 
this  text,  and  for  their  willingness  to  try  out  the  manuscript 
in  their  classes.  THE  AUTHORS 


CONTENTS 

PRELIMINARY  COURSE 


PAGE 

INTRODUCTION   .  1 


PART  I 

THE  STRAIGHT  LINE 6 

THE  ANGLE 14 

TRIANGLES 28 

POLYGONS 33 

THE  CIRCLE 36 

CIRCLE  AND  ANGLE 47 

PART   II 

AXIAL  SYMMETRY 53 

CONGRUENCE 61 

SURVEYING  .  65 


BOOK  I.    RECTILINEAR  FIGURES 

A  SYSTEM  OF  PROPOSITIONS 69 

ARRANGEMENT  OF  A  DEMONSTRATION 70 

AXIOMS.    PRELIMINARY  PROPOSITIONS 72 

CONGRUENCE 74 

CONSTRUCTIONS 84 

RIGHT  TRIANGLES 91 

PARALLEL  LINES 94 

ANGLE-SUM 102 

PARALLELOGRAMS     Ill 

THE  TRANSVERSAL  TIIKORKM.    TRAPEZOIDS 119 

INEQUALITIES 124 

COLLINEARITY  AND  CONCURRENCE 134 

ix 


x  CONTENTS 

BOOK  II.    THE  CIRCLE 

PAGE 

PRELIMINARY  PROPOSITIONS <,    .    .  143 

CHORDS  AND  SECANTS      145 

TANGENTS 153 

Two  CIRCLES 159 

ANGLE  MEASUREMENTS 164 

Loci 171 

COORDINATES 177 

CONSTRUCTIONS 181 

BOOK  III.    AREA 

PRELIMINARY  DEFINITIONS  AND  EXERCISES 193 

AREAS  OF  SIMPLE  FIGURES 199 

TRANSFORMATIONS 210 

THEOREM  OF  PYTHAGORAS 214 

BOOK  IV.    PROPORTIONAL  MAGNITUDES 
SIMILAR  POLYGONS 

REVIEW  OF  PROPORTION 231 

SIMILAR  POLYGONS.    INTRODUCTION 236 

PROPORTIONAL  SEGMENTS 240 

SIMILAR  TRIANGLES 244 

TRIGONOMETRIC  RATIOS 259 

CIRCLES  AND  PROPORTIONAL  LINES 265 

SIMILAR  POLYGONS 271 

NUMERICAL  PROPERTIES  OF  TRIANGLES 276 


BOOK  V.    REGULAR  POLYGONS  AND  CIRCLES 

CONSTRUCTION  OF  REGULAR  POLYGONS 285 

THEOREMS 293 

CIRCUMFERENCE  OF  A  CIRCLE.    PRELIMINARY  DISCUSSION      .    .    .  298 

AREA  OF  A  CIRCLE.    PRELIMINARY  DISCUSSION 305 

MENSURATION  OF  THE  CIRCLE.    FORMAL  DEMONSTRATION  ....  309 

MISCELLANEOUS  EXERCISES 319 

INDEX                                                                                                          .  327 


PLANE  GEOMETRY 


PRELIMINARY  COURSE 

1.  Origin  of  Geometry.  Egyptian  Geometry.  Geometry  is  one 
of  the  most  ancient  of  all  arts  and  sciences  It  arose  in  Baby- 
lonia and  Egypt  in  connection  with  such  practical  activities  as 
building,  surveying,  navigating,  etc.  In  fact,  the  word  "  geom- 
etry "  means  earth  measurement.  Herodotus,  a  Greek  histo- 
rian who  traveled  in  Egypt,  says  that  the  annual  overflowing  of 
the  Nile  changed  many  boundaries  in  the  adjoining  farm  land. 
Thus  it  became  necessary  to  measure  the  land  of  each  taxpayer 
every  year  in  order  that  taxes  might  be  properly  adjusted. 
In  this  way,  he  claims,  geometry  originated  in  Egypt,  and  all 
the  classical  writers  agree  with  him  in  calling  Egypt  the  home 
of  geometry. 

Much  new  light  was  thrown  on  early  geometry  by  the  dis- 
covery, in  recent  years,  of  Babylonian  inscriptions  and  Egyptian 


2       PLANE  GEOMETRY— PRELIMINARY  COURSE 

papyri.  As  early  as  1700  B.C.  an  Egyptian  scribe,  Ahmes,  wrote 
a  mathematical  treatise  containing  a  number  of  geometric  rules. 
Even  without  such  written  records  the  enormous  architec- 
tural works  of  the  ancients  —  their  pyramids,  obelisks,  temples, 
palaces,  and  canals  —  would  indicate  a  very  respectable  insight 
into  geometric  relations. 

2.  Greek  Geometry.  The  early  geometry,  however  valuable 
it  may  have  been  for  practical  purposes,  was  deficient  in  that 
it  consisted  simply  of  a  set  of  rules  obtained  after  centuries  of 


experimenting.  How  to  get  a  result  seemed  to  be  its  important 
question.  That  this  condition  did  not  last  indefinitely  is  due  to 
the  genius  of  the  Greeks.  Being  a  race  of  thinkers  and  poets, 
they  wished  to  know  .why  a  certain  result  must  follow  under  a 
given  set  of  conditions.  For  many  years  their  wise  men  studied 
in  Egypt.  Upon  returning  they  aroused  among  their  followers 
a  great  interest  in  the  study  of  geometry.  Many  new  truths  were 
now  discovered,  which  were  gradually  arranged  in  a  system. 
In  this  way  geometry  became  a  science.  After  about  three 
hundred  years  of  persistent  study  the  Greeks  produced  a  great 
masterpiece  in  the  form  of  Euclid's  "  Elements  of  Geometry  " 
(300  B.C.).  This  was  a  textbook  on  geometry,  divided  into  thir- 
teen chapters  or  "  books."  It  was  accepted  at  once  as  the  great 


PURPOSE  OF  GEOMETRY 


authority  on  all  geometric  questions,  and  has  retained  much  of 
its  importance  to  this  day. 

3.  Purpose  of  Geometry.    Geometry,  in  the  form  given  it  by 
the  Greeks,  is  no  longer  primarily  concerned  with  such  practi- 
cal activities  as  surveying.    Its  main  purpose  is  the  discovery 
and  classification  of  the  most  important  properties  of  points, 
lines,  surfaces,  and  solids,  in  their  relation  to  each  other. 

Hence  we  must  find  out  how  the  words  "  solid,"  "  surface," 
"  line,"  "  point,"  are  used  in  geometry. 

4.  Space,  Solids,  and  Surfaces.   Every  intelligent  being  has  a 
notion  of  what  space  is.    The  schoolroom  represents  a  por- 
tion of  space.    If  it 

were  not  bounded 
by  walls,  or  surfaces, 
it  would  extend  in- 
definitely. Therefore 
we  see  that  a  com- 
pletely inclosed  por- 
tion of  space  arises  only  when  there  exist  bounding  surfaces. 
Any  limited  portion  of  space  is  a  geometric  solid.  Solids  are 
bounded  by  surfaces. 

5.  Lines   and   Points.    Again,   the  walls  of   the   schoolroom 
would  extend    indefinitely   if    they   were    not   bounded.    But 
they  are    bounded  by  their   intersections,   the  edges   of   the 
room.  These  edges  are  lines.    Finally,  these  edges,  or  lines, 
would  extend  indefinitely  if  they  did  not  terminate  each  other 
by  their  intersections.    The  intersections  of  lines  are  points. 

6.  A  surface  may  be  considered  by  itself,  without  reference 
to  a  solid. 

A  line  may  be  considered  by  itself,  without  reference  to  a 
surface. 

A  point  may  be  considered  by  itself,  without  reference  to  a  line. 
Surfaces  are  either  plane  (flat)  or  curved. 
Lines  are  either  straight  or  curved. 


PLANE  GEOMETRY— PRELIMINARY  COURSE 


7.  Examples  of  Solids.  Geometric  solids,  surfaces,  lines,  and 
points  are  purely  ideal  objects.  An  ordinary  solid,  for  in- 
stance, is  material;  but  in  geometry  we  are  not  concerned 
with  the  matter  of  which  a  solid  is  composed.  We  are  inter- 
ested only  in  its  shape  and  size. 

The  following  diagrams  represent  some  of  the  most  common 
solids : 


CUBE 


CYLINDER 


CONE 


8.  Value  of  Geometry.  The  study  of  geometry  is  of  great 
value  for  two  reasons : 

In  the  first  place,  some  knowledge  of  geometric  principles  is 
indispensable  to  any  person  desiring  more  than  a  superficial 
acquaintance  with  engineering,  architecture,  designing,  draft- 
ing, physics,  astronomy,  surveying,  or  navigation.  Of  course 
it  is  impossible  in  an  elementary  book  to  point  out  in  detail 
the  connection  between  geometry  and  applied  science,  but  the 
fact  that  it  exists  would  be  a  sufficient  reason  for  the  appearance 
of  geometry  in  every  high-school  course. 

In  the  second  place,  there  is  no  other  subject  that  illus- 
trates more  clearly,  when  correctly  taught,  what  it  means  to 
prove  a  statement,  or  emphasizes  more  strongly  the  necessity 
of  accuracy  in  expression.  Geometry  also  greatly  increases 
one's  power  of  mathematical  reasoning,  secures  a  deeper  insight 


METHOD  OF  GEOMETRY  5 

into  spatial  relations,  and  gives  greater  skill  in  classification. 
If  to  this  be  added  the  fact  that  the  study  of  geometry  re- 
quires no  special  mental  equipment  beyond  ordinary  common 
sense,  and  that  geometric  truths  once  established  hold  for  all 
time,  no  further  explanation  is  needed  for  the  interest  shown 
by  many  great  thinkers  in  geometric  investigations,  nor  for 
the  prevailing  opinion  that  every  educated  person  ought  to 
know  something  of  geometric  methods. 

9.  Method  of  Geometry.  The  history  of  geometry  shows  that 
whenever  the  subject  was  studied  merely  for  practical  pur- 
poses few  important  discoveries  were  made.  The  validity  of 
a  principle  was  determined  by  experiment,  and  people  were 
satisfied  if  they  found  how  to  get  a  certain  result.  Progress 
began  as  soon  as  the  attempt  was  made  to  give  reasons  for 
processes  and  rules.  At  the  present  time  geometry  can  be 
studied  either  by  experimenting  or  by  reasoning.  Either 
method  is  valuable,  but  it  requires  little  reflection  to  see  that 
a  person  who  knows  why  he  proceeds  in  a  certain  way  is  on 
safer  ground  than  one  who  follows  a  rule  blindly.  The  second 
method  is,  however,  the  more  difficult  one.  In  order  to  lessen 
the  difficulty  this  book  has  been  divided  into  two  parts,  an 
informal  part  and  a  formal  part.  The  introductory  course  will 
lead  by  slow  degrees  from  one  method  to  the  other.  At  first 
simple  observational  tests  will  be  used;  later  a  reason  will 
be  required  for  every  statement. 


PART  I 

THE  STRAIGHT  LINE 

10.  General  Properties.    The  straight  line  and  the  plane  are 

among  the  most  fundamental  concepts  of  geometry.  For  that 
reason  it  is  difficult,  if  not  impossible,  to  define  them.  The 
following  simple  tests  may  take  the  place  of  definitions  : 

1.  Lay  your  pencil  on  the  cover  of  your  book.    The  pencil 
may  be  moved  about  freely  on  the  surface  of  the  cover,  how- 
ever you  happen  to  place  it.    Could  this  be  done  on  a  curved 
surface  ? 

2.  Fold  a  sheet  of  paper.  The  crease  represents  a  straight  line. 

3.  If  an  elastic  band  or  a  piece  of  thread  is  stretched,  it 
assumes  the  appearance  of  a  straight  line. 

4.  Look  along  the  edge  of  your  ruler.   If  the  edge  is  straight, 
it  can  take  on  the  appearance  of  a  point. 

EXERCISES 

1.  Locate  on  your  paper,  near  the  top,  any  convenient  point. 
Represent  this  point  by  a  pencil  dot.    Name  the  dot  A. 

2.  Through  this  point  draw  a  straight  line.    Observe  that  the 
drawing  is  not  really  a  line,  since  lines  are  ideal  objects,  but  it 
serves  to  represent  a  line. 

3.  How  would  you  test  the  straight-ness 
of  this  line  ? 


________________  jr_J_  ___ 

5.  Draw  a  dotted  line  through  A  ;  a  spaced  line  (see  figure). 

6.  How  many  straight  lines  can  be  drawn  through  a  given 
point  ? 


LINE-SEGMENTS  7 

7.  How  many  points  can  two  straight  lines  have  in  common  ? 

8.  Explain  why  a  point  is  sometimes  represented  by  the 
symbol  X. 

9.  Draw  a  straight  line  through     ~~* * 

the  points  A  and  B. 

10.  How  many  straight  lines  can  be  drawn  through  these 
two  points  ? 

11.  Of  what  value  is  this  fact  in  testing  the  straightness  of 
a  ruler  ?  in  testing  a  plane  surface  ?  in  supporting  a  rod  ? 

12.  Into  how  many  parts  is  a  straight  line  divided  by  two 
points  on  it  ? 

11.  Summary.    Properties  of  Straight  Lines  inferred  directly: 

1.  Through  a  given  point  an  indefinite  number  of  straight 
lines  may  be  drawn. 

2.  Tivo  straight  lines  can  intersect  in  but  one  point. 

3.  Only  one  straight  line  can  be  draivn  through  two  given  points. 

4.  Two  straight  lines  that  have  two  points  in  common  coincide 
throughout  and  form  but  one  line. 

5.  Two  straight  lines  do  not  inclose  a  space. 

12.  Line-Segments.    Measurement  of  Segments.    Any  drawing 
of  a  straight  line  in  reality  represents  but  a  small  portion  of 
that  line.    Our  imagination,  however,  can  extend  this  limited 
portion  indefinitely.    It  is  sometimes  convenient  to  distinguish 
between  a  limited  and  an  unlimited  line. 

Any  portion  of  a  straight  line  lying  between  two  of  its 
points  is  called  a  segment  of  that  line.  A  B 

A  straight  line  may  be  designated  by  ~ 

naming   two    of   its    points,    as    the  line 
AB]  or  by  a  single  (small)  letter,  as  the  Q  D 

line  a. 


I I 

A  segment  is  designated  by  naming  its 

end  points  (extremities),  as  the  segment  CD;  or  by  a  single 
(small)  letter,  as  the  segment  m. 


8       PLANE  GEOMETRY— PRELIMINARY  COUKSE 


EXERCISES 

1.  How  many  line-segments  are  there  in  the  capital  letter  M  ? 
Indicate  each  segment  by  naming  its  extremities. 

2.  How  would  you  test  the  equality  or  inequality  of  two 
segments  ? 

3.  Draw  two  straight  lines.   Indicate  on  the  first  an  arbitrary 
segment  a.  On  the  second  line  mark  off  a  a 
segment  equal  to  a  by  using  (1)  a  strip 

of  paper  having  a  straight  edge;  (2)  a      

ruler ;  (3)  a  pair  of  dividers.    Which  of  these  methods  is  the 
most  convenient  ? 

4.  Draw  five  segments  of  different  lengths.    How  can  you 
compare  one  of  them  with  a  given  length  on  the  scale  ?    Why 
do  we  use  a  standard  scale  ?   Name  the  principal  standards  of 
length. 

5.  Measure   the   five  segments  in  inches;    in  centimeters. 
Tabulate  your  readings  as  follows: 


Segment 

Inches 

Centimeters 

a 

b 

c 

d 

e 

6.  Draw  segments    of  the   following  lengths :   2  in.,  2^  in., 
5J-  in.,  4y\  in.,  12  cm.,  8.8  cm. 

7.  Draw  a  segment  10  cm.  long  and  measure  its  length  in 
inches.    How  many  inches  are  there  in  a  centimeter?    Test 
your  result  by  taking  off,  with  the  dividers,  one  centimeter  on 
the  centimeter  scale  and  applying  it  to  the  inch  scale. 

8.  Draw  a  segment  5  in.  long  and  find  its  length  in  centi- 
meters.   From  this  calculate   the  number  of  centimeters  in 
1  inch. 


OPPOSITE  DIRECTIONS  9 

9.  Mark  off  on  a  line  ten  consecutive  divisions  of  your 
centimeter  scale.   How  can  you  test  the  accuracy  of  the  scale  ? 

10.  Draw  on  your  paper,  or  on  the  blackboard,  several  seg- 
ments.   Estimate  their  lengths  and  test  by  measuring  them. 

11.  Estimate  the  comparative  lengths 

of  the  segments  a  and  b  in  the  figure,       x> <\ 

and  then  test  your  estimate  by  means  ^        b      x^ 

of  the  dividers. 

12.  On  a  map  measure  in  inches  or  centimeters  the  dis- 
tances between  four  cities  chosen  at  random.    With  the  aid 
of  the  scale  given  on  the  map  express  these  distances  in  miles 
or  kilometers. 

13.  Can  you  measure  with  absolute  accuracy  ?    If  not,  why 
are  different  degrees  of  accuracy  possible  ? 

13.  Summary.    1.  A  segment  is  measured  by  comparing  it 
with    given    standards    of   length.    Measurements    are    only 
approximate.    The  degree  of  accuracy  obtained  will  depend 
chiefly  on  the  size  of  the  unit  chosen ;  that  is,  the  smaller  the 
unit,  the  more  correct  will  be  the  measurement. 

2.  If  the  extremities  of  a  segment  are  given,  the  entire  segment 
is  thereby  determined  both  as  to  position  and  length. 

3.  The  segment  joining  two  points  is  known  as  the  measure 
of  the  distance  between  the  points. 

4.  Two  segments  whose  extremities  can  be  made  to  coincide 
must  coincide  throughout. 

5.  Two  segments  a  and  b  must  be  in  one  of  three  relations  to 
each  other : 

a  >  b  (a  is  greater  than  b), 

a  =  b  (a  equals  b), 

a  <C  b  (a  is  less  than  &). 

14.  Opposite  Directions.  The  Fundamental  Operations.  A  satis- 
factory understanding  of  some  new  terms  and  processes  may 
best  be  gained  by  solving  the  following  exercises : 


10      PLANE  GEOMETEY— PEELIMINAEY  COUESE 

EXERCISES 

1.  Given  a  line  I  and  a  point  A  on  it.    Eequired  to  change 
the  position  of  A  on  I  by  2  cm.    How  many  solutions  are  pos- 
sible ?    On  your  drawing  denote  the 

new  positions  of  A   by  Al  and  A2      I 

(read  "  A  one  "  and  "  A  two  ").  Z 

2.  How  many  directions  can  be  distinguished  on  a  straight 
line  from  a  given  point  on  that  line  ? 

3.  Prolong  a  segment  AB  =  4  cm.  by  3  cm.  in  the  direction 
AB ;  in  the  direction  BA. 

4.  How  many  directions  are  indicated 
by  two  straight  lines  crossing  each  other  ? 
by  three  lines  passing  through  one  point  ? 
four  lines  ?    n  lines  ? 

5.  Draw  a  segment  about  6  in.   long 

and  on  it  take  off  in  succession  AB  =  2  in.,  £C  =  If  in., 
CD  =  If  in.  Find  the  length  of  AD  by  computation  and  by 
measurement. 

6.  A  man  walks  4.5  mi.  due  north,  and  then  3  j  mi.  due  south. 
How  far  is  he  from  the  starting  point  ?   Draw  a  diagram  of 
his  route,  1  mi.  being  represented  by  1  in. 

7.  A  boat  sails  due  east  for  3  hr.,  at  the  rate  of  7  mi.  an  hour. 
It  then  returns  over  the  same  course  at  the  rate  of  6.5  mi.  an 
hour.  Ascertain  by  means  of  a  drawing  how  far  the  boat  is  from 
the  starting  point  2  hr.  after  turning  about.   (Eepresent  1  mi. 
by  1  cm.) 

8.  On  a  straight  line  lay  off,  four  times  in  succession,  a 
segment  2.5  cm.  long.    Measure  the  resulting  segment.    Test 
the  result  by  calculation.    How  might  the  dividers  be  used  to 
advantage  in  this  exercise  ? 

9.  Divide  a  line  15  cm.  long  into  five  equal  parts  by  inspec- 
tion ;  by  measurement. 


FUNDAMENTAL  OPEBATIONS  11 

10.  On  a  straight  line  locate  in  order  the  points,  A,  B,  C,  D, 
E,  F.    Express 

A  C,  BD,  CF,  each  as  the  sum  of  two  segments ; 
AD,  BE,  each  as  the  sum  of  three  segments ; 
AB,  BD,  each  as  the  difference  of  two  segments ; 
AB  +  BD,   BD  +  DF,    AB  +  BD  +  DE,    each   as    a   single 
segment. 

11.  A,  B,  C,  are  points  on  a  line.    If  AB  =  BC,  how  is  B 

situated  ? 

12.  Draw  a  line-segment  and  locate  its  middle  point  by 
inspection.    Test  by  measurement. 

15.  Summary.    The  preceding   exercises  justify  the  state- 
ments of  this  section  and  the  two  following. 

Any  point  of  a  straight  line  divides  it  into  two  parts,  which 
are  said  to  lie  on  opposite  sides  of  that  point,  or  to  have  opposite 
directions.    Each  of  these  parts  is 
sometimes  called  a  ray.  4 ? £ 

The  point  from  which  a  ray  ex- 
tends is  called  its  origin.    Thus,  in  the  figure,  OA  and  OB  are 
rays  having  the  common  origin  0. 

16.  Segments  'may  be  added  together.     Of  two  unequal  seg- 
ments the  smaller  may  be  subtracted  from  the  larger.    Segments 
may  be  multiplied  or  divided  *  by  a  number. 

17.  If  on  a  line  three  points,  A,  B,  C,  are  taken  in  succes- 
sion, so  that  AB  =  BC,  B  is  called  the  mid-point  of  segment  A  C, 
or  B  is  said  to  bisect  segment  AC,  and  A  and  C  are  equally 
distant  from  B. 

18.  Equality  of  Segments.    A  number  of  fundamental  alge- 
braic processes  are  also  of  great  importance  in  geometry,  as 
the  following  exercises  will  suggest. 

*  Division,  at  this  point,  is  possible  only  if  the  given  segment  can  be 
measured. 


12     PLANE  GEOMETRY— PRELIMINARY  COURSE 

EXERCISES 

1.  Given  a  segment  a.    Draw  a  segment  equal  to  3  a. 

2.  Given  two  segments  a  and  b.    Draw  segments  equal  to 
a  -f-  b}  a  —  b,  3  (a  —  b),  taking  a  >  b. 

_  7- 

-    „' 


o.   jju  -fig.  J-,                    M.  —  a  j 
and                                     b  =  b'. 

&'               b 

Explain  why         a  -f-  £  =  a'  -f-  b'. 

4.  If,  in  Fig.  2,  AC  =  A'C',  and 
AB  —  A  'B',  explain  why  EC  —  B'C'. 

FIG.  1* 

A                   B             C 

i                     I               i 

5.  In  Fig.  3,  AB  =  CD.  Find  two 
other  equal  segments. 

6.  Given  AB  =  3  a,  and  CD  =  3b 

(Fig.  4).     If    a  =  b,    explain    why 
AB  =  CD. 

7.  In  the  same  figure 
AB             CD                 ^jL_ 

A'                   B'            C' 

I                      i               i 

FIG.  2 

A         B                CD 

II                   II 

1            1                   1            I 
FIG.  3 

a               a               a      B 

33                    c 

6    i    b    ,    6   f 

If   .4j3  =  G\D,    explain  why 

FIG.  4 

8.  Express  in  words  the  principles  underlying  Exs.  3~7. 


REVIEW  EXERCISES 

1.  Locate  on  a  straight  line,  successively,  one  point ;  two  points ; 
three  points;    n  points.    How  many  segments  arise  in  each  case? 
How  many  rays  ? 

2.  What  is  the  largest  number  of  points  in  which  three  lines  can 
intersect  ?  four  lines  ?  five  lines  ?  n  lines  ?   Arrange  your  answers  in 
the  form  of  a  table. 

3.  What  is  the  largest  number  of  lines  determined  (fixed)  by  two 
points  ?  three  points  ?  n  points  ? 

*  a'  is  read  "  a  prime." 


KEVIEW  EXERCISES 


13 


4.  Given  the  segments  p  =  28  mm.,  q  —  16  mm.,  r  =  9  mm.,  s  = 
37  mm.    Draw  segments  equal  top  +  </ ;  s  +  r ;  s  —  q;  p  —  r ;  p  +  r+cy; 
p  +  ,7  -  s.  4 

5.  Given  a  =  15  cm.,  &  =  6  cm.    Draw  segments  equal  tB  a  "*"    • 
ft     2a      3ft 

-3;T  +  T" 

6.  A  segment  12  cm.  long  is  to  be  divided  into  three  parts  which 
are  to  be  in  the  ratio  of  1,  2,  3.    Find  their  lengths. 

Solution.  1  +  2  +  3  =  6. 

12  -*•  6  =  2. 

1x2  =  2;  2x2  =  4;  3x2  =  6. 
Check.  2  +  4  +  6  =  12. 

7.  The  sum  of  the  length  and  the  width  of  a  house  is  70  ft.    The 
width  is  to  the  length  as  3  is  to  4.  What  are  the  dimensions  of  the 
house  ? 

8.  A  segment  x  units  long  is  to  be  divided  into  five  parts  which 
are  to  be  in  the  ratio  of  1,  2,  3,  4,  5.    Determine  the  length  of  each 
part. 

9.  A  segment  18  in.  long  is  divided  into  three  parts  such  that 
the  first  is  equal  to  the  sum  of  the  other  two,  while  the  sum  of  the 
first  two  is  eight  times  the  third.    Find  the  length  of  each  part. 

10.  From  a  fixed  point  draw  five  rays.  Lay  off  on  these  rays,  from 
the  common  origin,  the  segments  a,  ft,  c,  cl,  e,  respectively,  making 
a  =  2  cm.,  ft  =  3  cm.,  c  =  4  cm.,  d  =  5  cm.,  e  =  6  cm.  Join  the  ex- 
tremities of  these  segments  in  order,  naming  the  new  segments 
DI,  n,  p,  q.  Measure  m,n,p,  q,  and  tabulate  your  results  as  follows: 


Segment 

Estimated 
length 

Measured 
length 

Actual 
error 

Percentage 
error 

m 

n 

P 

11.  Will  the  values  of  m,  n,  />,  and  q  be  the  same  in  each  case  for 
various  figures  drawn  in  accordance  with  the  above  directions  ? 


14  PLANE  GEOMETRY— PRELIMINARY  COURSE 


THE  ANGLE 

19.  Definition  and  Notation.    An  angle  is  a  figure  formed  by 
two  rays  having  a  common  origin.    The  two  rays,  AB  and  AC, 
are  called  the  sides,  and  A,  their  origin, 
is  called  the  vertex,  of  the  angle. 

The  outstretched  fingers,  the  hands  of  a  clock, 
the  spokes  of  a  wheel,  the  divisions  of  the  com- 
pass,  the  gable  of  a  roof,  and  many  similar  illus- 
trations readily  suggest  the  frequent  occurrence  and  great  importance 
of  angles. 

The  methods  of  naming  angles  are  shown  in  the  figures 
below. 


A  single  angle  at  a  point  may  be  designated  by  a  capital 
letter  placed  at  its  vertex,  as  angle  0. 

When  several  angles  have  the  same  vertex,  each  angle  may 
be  designated  by  three  letters ;  namely,  by  one  letter  on  each 
of  its  sides,  together  with  one  at  its  vertex.  The  letter  at  the 
vertex  is  read  between  the  other  two,  as 
angle  DA  C,  angle  CAB,  etc. 

Sometimes  an  angle  is  denoted  by  a 
small  letter  placed  between  its  sides  near 
the  vertex,  as  angle  m. 

The  sides  may  be  denoted  by  small 

letters.  The  angle  ab  is  the  angle  formed  by  the  rays  a  and  b. 
This  notation  is,  however,  used  less  frequently  than  the 
preceding. 

The  word   "angle"  is  frequently  replaced  by  the  symbol  /..   The 
symbol  d  is  used  as  an  abbreviation  for  "  angles." 


ANGLES  15 

EXERCISES 

1.  How  many  angles  may  be  found  in  the  letters  N  and  W  ? 

2.  From  a  given  point  draw  three  rays  ;  four  rays  ;  five  rays. 
How  many  angles  are  formed  in  each  case  ? 

3.  Through  a  given  point  draw  two  lines  ;  three  lines  ;  four 
lines.    How  many  angles  are  formed  in  each  case  ? 

4.  Draw  two  angles,  making  them  as  nearly  equal  as  you 
can.   Test  their  equality  by  copying  one  of  the  angles  on  tracing 
paper,  or  by  cutting  out  one  of  the  angles,  and  applying  it  to 
the  other. 

5.  The  figure  shows  how  a  strip  of  paper  having  a  straight 
edge  may  be  used  to  copy  an  angle. 

Place  the  straight  edge  on  the  side 
AB  of  Z.  BA  C,  and  mark  on  it  the 
position  of  the  vertex  A.  Mark 
also  the  point  D,  where  the  side 
AC  intersects  the  other  edge  of 
the  paper.  Then  Z.DAE=^CAB. 
With  the  aid  of  this  paper  strip  draw  an  angle  equal  to  Z  CAB. 

6.  If  you  extend  the  sides  of  Z.CAB  beyond  C  and  B,  do 
you  change  the  size  of  the  angle  ? 

7.  Draw  a  ray  and  place  your  pencil  on  it.    Revolve  the 
pencil  on  the  flat  surface  of  the  paper,  using  one  extremity  of 
the  pencil  as  a  pivot.    Observe  that  each  position  of  the  pencil 
indicates  a  different  angle.    The  pencil  will  eventually  return 
to  its  first  position.  This 

rotation  evidently  gives 
a  picture  of  every  pos- 
sible angle. 

8.  Show  that   in   the  b  b 
figures  a  line  could  revolve  from  the  position  a  to  the  posi- 
tion b,  or  from  I  to  a. 

This  fact  is  indicated  by  the  arrowheads. 


16     PLANE  GEOMETRY— PRELIMINARY  COURSE 


0 


CLASSIFICATION  OF  ANGLES 

(A)  THE  THREE  FUNDAMENTAL  ANGLES 

20.  From  the  preceding  exercises  may  be  derived  a  general 
idea  of  the  magnitude  of  an  angle.    It  appears  that  the  magni- 
tude of  an  angle  depends  on  the  amount  of  revolution  necessary 
to  turn  a  line  through  the  angle  about  the  vertex  as  a  pivot. 
The  revolving  line  is  said  to  generate  or  describe  the  angle.    The 
size  of  an  angle  is  independent  of  the  length  of  its  sides. 

21 .  A  round  angle,  or  perigon,  is  an  angle 
whose  magnitude  is  indicated  by  a  com- 
plete revolution  of  the  generating  line. 

22.  All  round  angles  are  equal. 

23.  A  straight  angle  is  an  angle 
whose  sides  are  in  the  same  straight      ^ 
line,  on  opposite  sides  of  the  vertex ; 

as  Z.  A  OB. 

24.  All  straight  angles  are  equal.    Why  ? 

25.  Two  angles  having  the  same  vertex 
and  a  common  side  between  them  are  called 
adjacent  angles;  as  the  AAOB  and  BOC.  O 

EXERCISES 

1.  Can  two  angles  have  a  common  side  and  a  common  ver- 
tex without  being  adjacent  ? 

2.  In  the  annexed  figure  Z.  A  OC 
is  a  straight  angle.    If  a  ray  OB 
revolves    from    the    position    OA 
toward  the  position  OC,  two  angles    Q — 
are   formed,    Z.m    and   Z-n.     At 

first  Z.  m  is  less  than  Z.  n.  Finally,  however,  /.  m  will  be 
greater  than  Z.  n.  Hence  one  position  of  OB  must  have  made 
the  angles  equal.  Indicate  approximately  this  position  of  OB 
in  the  figure. 


•B 


n/m 
O 


ANGLES 


17 


3.  Take  a  piece  of  paper  having  a  straight  edge  AB  and 
fold  it  so  as  to  bring  the  point  B  on  the  point  A.  Open  the 
paper  and  name  the  crease  OC.  Then  /LAOC  =  Z.BOC. 


26.  A  right  angle  is  an  angle  such  that  the  adjacent  angle 
formed  by  extending  one  of  its  sides  beyond  the  vertex  is 
equal  to  it.    Thus,  in  the  figure,  Z.BOC  is  a  right  angle. 

27.  A  right  angle  is  half  of  a  straight  angle.    Why  ? 

28.  All  right  angles  are  equal.    Why  ? 

29.  The  sides  of  a  right  angle  are  said  to  be  perpendicular  to 
each  other;  thus  OC  in  the  figure  is  perpendicular  to  OB,  and 
OB  is  perpendicular  to  OC.    This  is  sometimes  abbreviated  as 
follows :   OC  _L  OB,   OB±OC.    If  OC  is  perpendicular  to  AB, 
0  is  called  the  foot  of  the  perpendicular  OC. 

30.  Eound  angles,  straight  angles,  and  right  angles  are  so 
important  that  they  are  taken  as  standards  with  which  other 
angles  are  compared. 

31.  If  three  rays,  a,  b,  and  c,  are  drawn  from  the  same  point, 
as  shown  in  the  accompanying  figure,  the  following  relations 
exist : 

1.  /-  ac  is  the  sum  of  Z  ab  and  /-  be. 

2.  /.ah  is  the  difference  between  /-ac 
and  Z  be. 

3.  Zac  is  greater  than  either  /.ab  or  /-be. 

4.  If  Z  ab  =  Z  be,  the  ray  b  is  the  bisector 
of  Z  ac. 

32.  Tivo  angles  A  and  B  must  be  in  one  of  the  following  three 
relations  to  each  other:  Z.A>/.B,  Z.A  =  /LB,  or  /-A  <Z.B. 


18     PLANE  GEOMETEY— PEELIMINAEY  COUESE 


(B)    THE   THREE   OBLIQUE   ANGLES 

33.  An  acute  angle  is  an  angle  less  than  a  right  angle,  as  the 
angle  COD. 

34.  An  obtuse  angle  is  an  angle  greater  than  a  right  angle 
but  less  than  a  straight  angle,  as  angle  AOB. 

35.  A  reflex  angle  is  an  angle  greater  than  a  straight  angle 
but  less  than  a  round  angle,  as  angle  MON. 


B 


D 


36.  Acute,  obtuse,  and  reflex  angles  are  sometimes  called 
oblique  angles,  and  intersecting  lines  not  mutually  perpendicular 
are  said  to  be  oblique  to  each  other. 


EXERCISES 


1.  What  kind  of  angle  is  each  of  the  following  angles  of 
the  annexed  figure :  Z  A  OB, 
Z  A  OD,    Z  A  OE,    Z  A  OF, 
Z  BOD,     Z  BOE,     Z  BOF, 


2.  In    the    same    figure 
what  is  the  relation  of  OA 
to  OD?   OC  to  OD?   OB  to 
OE?   OB  to  OF? 

3.  Draw  an  acute  angle  ; 
a  right  angle  (using  a  paper 
pattern   obtained   by  fold- 
ing) ;  an  obtuse  angle. 

4.  Point  out  right  angles  in  the  construction  and  equipment 
of  the  schoolroom  ;  in  the  exterior  of  a  house  j  in  the  street, 


ANGLE-PAIRS 


19 


(C)   THE   FOUR    SPECIALLY   RELATED   ANGLE-PAIRS 

37.  Two  angles  whose  sum  is  equal  to  a  right  angle  are 
said  to  be  complementary.    Each   of  the  angles  is  called  the 
complement  of  the  other. 
D 


Thus  Z  BOG  is  the  complement  of  Z  COD  •  also  Z  a  is  the  complement 
of  Zb. 

38.  Two  angles  whose  sum  is  equal 
to  a  straight  angle  are  said  to  be  supple- 
mentary, each  being  the  supplement  of 
the  other. 

In  the  accompanying  figure  A  AOC  and     A — 
BOG  are  supplementary. 

39.  Two  angles  whose  sum  is  a  round  angle  are  said  to  be 
conjugate  angles. 

40.  If  two  lines,  AB  and  CD,  intersect  at 
0,  as  in  the  figure,  the  A  AOC  and  BOD  are 
called  vertical  or  opposite  angles ;  also  the  A 
AOD  and  BOC. 

EXERCISES 

1.  Draw  an  acute  angle,  and  then  (using  a  pattern  of  a  right 
angle)  draw  its  adjacent  complement.    This  may  be  done  in 
two  ways.    Explain. 

2.  Draw  an  angle  and  construct  its  adjacent  supplement. 
This  may  be  done  in  two  ways.    Explain. 

3.  Draw  the  supplement  and  the  complement  of  a  given 
angle.    Show  that  the  difference  of  these  two  angles  is  a  right 
angle. 


20      PLANE  GEOMETRY— PRELIMINARY  COURSE 

4.  If   two  angles  are  equal,  how   do   their  complements 
compare  ?  their  supplements  ?    Why  ? 

5.  What  kind  of  angle  is  equal  to  its  supplement  ?  less 
than  its  supplement  ?    greater  than  its  supplement  ? 

6.  If  one  of  two  rods  is  made  to  turn  about  a  point  in  the 
other  as  a  pivot,  two  pairs  of  vertical  angles  are  formed.    As 
the  vertical  angles  are  generated  by  the  same  amount  of  turn- 
ing, how  do  they  compare  in  size  ? 

7.  Show  the  equality  of   two  vertical  angles  by  copying 
one  of  them  on  tracing  paper. 

8.  Given  an  angle  a.    Construct  its 
adjacent    supplement    in    two    ways. 
Does  this  suggest  another  reason  for 
the  equality  of  vertical  angles  ? 

9.  In  the  accompanying  figure  find  the  vertical  angle  of 
Z  a  -\-  Z.  I ;    the   supplement   of   Z.  I ;    the 

supplement  of  Z.  b  +  Z.  d.  How  many  pairs 
of  vertical  angles  and  of  supplementary 
adjacent  angles  in  the  figure  ?  What  is 
the  sum  of  Z.  a,  Z.  c,  and  Z.  e  ? 

10.  Prove  that  the  vertical  angles  of  two 
complementary  angles  are  also  complemen- 
tary, and  that  the  vertical  angles  of  two  supplementary  angles 
are  supplementary. 

11.  State  the  conclusions  obtained  in  each  of  the  following 
cases : 

1.  Z.  a  is  the  supplement  of  Z.  b,  and 
Z.  b  is  the  supplement  of  Z.  c. 

2.  Z.  a  is  the  complement  of  /-  b,  and 
Z.  b  is  the  complement  of  Z  c. 

3.  Z  a  is  the  supplement  of  Z  b,  and 
Z.  b  is  the  complement  of  /-  c. 

4.  Z  a  is  the  supplement  of  Z  b,  Z  b  =  Z  c,  and 
Z  c  is  the  supplement  of  Z  d. 


SUMMARY  21 

5.  The  supplement  of  Z-a  equals  the  supplement  of  Z.b. 

6.  The  supplement  of  Z  a  is  greater  than 
the  supplement  of  Z  b. 

12.  Show   that    the    supplements    of   two  \c/ 

complementary  angles  are  together  equal  to  J^ 

three  right  angles.  / 

13.  Through  the  vertex  of  a  right  angle     ' 
draw  a  straight  line  falling  with- 
out the  angle.    Prove  that  the  two 

acute  angles  formed  are  comple- 
mentary. 

14.  In    the   figure  Z  a   is   the 
supplement  of  Z  c.    Prove  that 
Z  b  is  the  supplement  of  Z  d. 

15.  If  two  angles  are  comple- 
mentary, what  is  the  relation  of 

their  complements  ?    If  two  angles  are  supplementary,  what 
is  the  relation  of  their  supplements  ? 

41.  Summary.    The  truth  of   the  following  statements  is 
now  apparent : 

1.  All  round  angles  are  equal. 

2.  All  straight  angles  are  equal. 

3.  All  right  angles  are  equal. 

4.  At  a  given  point  in  a  given  line  only  one  perpendicular 
can  be  drawn  to  that  line  (in  the  same  plane). 

5.  The  complements  of  the  same  angle,  or  of  equal  angles, 
are  equal. 

6.  The  supplements  of  the  same  angle,  or  of  equal  angles, 
are  equal. 

7.  Vertical  angles  are  equal. 

8.  If  two  adjacent  angles  have  their  exterior  sides  in  a  straight 
line,  they  are  supplementary. 

9.  If  two  adjacent  angles  are  supfnementary ,  their  exterior 
sides  lie  in  a  straight  line. 


22     PLANE  GEOMETRY— PRELIMINARY  COURSE 


6 


MEASUREMENT  OF  ANGLES 

42.  From  the  preceding  study  of  angles  it  is  evident  that  a 
round  angle  equals  two  straight  angles  or  four  right  angles. 

Fold  a  piece  of  paper  of  any  shape  and  call  the 
straight  folded  edge  AB.  Fold  again  so  as  to  bring 
B  on  A.  Now  open  the  paper.  The  four  angles 
formed  by  the  creases  are  right  angles.  Why? 
By  continuing  to  fold  the  paper  properly  we  may 
obtain  eight  equal  angles,  sixteen  equal  angles,  etc. 
This  is  a  mechanical  way  of  dividing  the  round 
angle  into  4,  8,  16,  etc.,  equal  parts.  D 

43.  In  measuring  angles  we  imagine  the  angular  magnitude 
about  a  point  in  a  plane  to  be  divided  into  360  equal  parts, 
called  degrees,   and  we  simply  state  how  many  degrees  the 
angle  under  consideration  contains.    Thus  the  degree  is  ^^ 
of  a  round  angle,  ^\^  of  a  straight  angle,  -fa  of  a  right  angle. 

The  degree  is  again  divided  into  60  equal  parts  called 
minutes,  and  the  minute  into  60  equal  parts  called  seconds. 
The  expression  40  degrees  7  minutes  20  seconds  is  written 
40°  7 '  20". 

44.  A  special  instrument,  called  a  protractor,  is  frequently  used  for 
angle  measurements.   The  figure  represents  one  form  of  the  protractor. 
By  joining  the  notch  0  of  the 

protractor  to  each  graduation 
mark  we  obtain  a  set  of  angles 
at  0,  each  usually  represent- 
ing an  angle  of  one  degree. 

To  measure  a  given  angle 
with  the  protractor,  place  the 
notch  of  the  protractor  at 
the  vertex  of  the  angle,  and  the 
base  line  along  one  side  of 

the  angle.   The  other  side  of  the  angle  then  indicates  on  the  protractor 
the  number  of  degrees  in  the  angle. 

To  draw  an  angle  of  a  given  number  of  degrees,  place  the  base  of  the 
protractor  along  a  straight  line  and  mark  on  the  line  the  position  of  the 
notch  0.  Then  place  the  pencil  at  the  required  graduation  mark  and 
(after  removing  the  protractor)  join  the  point  so  marked  to  0. 


PRELIMINARY  COURSE 


EXERCISES 


23 


1.  Divide  a  straight  angle  into  three  parts,  a,  ft,  c.  Estimate  the 
size  of  these  angles.  Then  measure  each  angle  with  the  protractor 
and  tabulate  your  results  as  follows : 


DEGREES 

A  vr*  T  TT 

Estimated 

Measured 

a 

b 

c 

Sum 

i 

2.  Divide  a  round  angle  into  five  parts  and  proceed  as  in  Ex.  1. 

3.  With  a  protractor  draw  angles  of  20°,  30°,  45°,  90°,  36°,  120°, 
135°,  225°. 

4.  How  many  degrees  in  the  following  fractions  of  a  right  angle  : 
?>  i>  i>  i>  s>  4?  iV?  jV>  rV>  TJU>  32(J>  f >  f  >  f >  f >  i?  f\r  ? 

5.  How  many  degrees  in  the  following  fractions  of  a  straight 
angle :  £,  J,  J,  i,  J,  J,  J,  T\>,  ^,  TV,  y1*,  ^  3o>  sV>  ?V>  A  ? 

6.  Which  integers  are  exactly  contained  in  360  ?   Does  this  sug- 
gest why  a  round  angle  is  divided  into  360  parts  ? 

7.  How  many  degrees  in  the  complement  of  each  of  the  follow- 
ing angles :  20°,  30°,  45°,  67°,  89°,  a°,  (a  +  ft)0,  (45  +  a)°? 

8.  How  many  degrees  in  the  supplement  of  each  of  the  following 
angles :  60°,  45°,  90°,  110°,  x°,  (x  -  a)°? 

9.  An  angle  is  ^  (^,  |)  *  of  its  supplement.    How  many  degrees 
in  the  angle? 

10.  An  angle  is  3  (2,  4,  5)  *  times  as  great  as  its  complement.  How 
many  degrees  does  it  contain  ? 

11.  The  supplement  of  an  angle  is  three  times  its  complement. 
How  many  degrees  in  the  angle  ? 

Suggestion.   180  —  x  =  3  (90  —  x). 

*  Each  of  the  values  given  in  the  parenthesis  is  to  be  substituted  in 
place  of  the  value  given  before  the  parenthesis,  making  a  new  exercise. 


24     PLANE  GEOMETRY— PRELIMINARY  COURSE 

12.  What  fractions  of  a  right  angle  are  the  angles  formed  by 
the  hands  of  a  clock,  from  hour  to  hour,  between  12  and  6  o'clock  ? 
What  fractions  of  a  straight  angle?    How  many  degrees  in  each 
angle  ?  State  in  each  case  whether  the  angle  is  acute,  right,  or  obtuse. 

13.  How  many  degrees  in  the  angles  formed  by  the  hands  of  a 
clock  when  the  time  is  3.12,  11.48,  10.48,  8.24,  9.36,  6.20,  7.22 
(railroad   notation)  ?    (Remember   that   the   minute   hand   travels 
12  times  as  fast  as  the  hour  hand,  and  that  a  minute  division  on 
the  face  of  the  clock  represents  6°.) 

14.  In  how  many  minutes  does  the  minute  hand  revolve  through 
an  angle  of  90°?  45°?  270°?   In  how  many  hours  does  the  hour  hand 
describe  these  angles? 

15.  What  angle  is  described  in  an  hour  by  the  minute  hand?   by 
the  hour  hand  ? 

16.  A  man,  setting  his  watch,  moves  the  minute  hand  forward 
half  an  hour  and  then  moves  it  back  8  min.    How  many  degrees  in 
the  angle  between  the  first  and  the  final  position  of  the  minute 
hand? 

17.  The  driving  wheel  of  an  engine  revolves  10  (20,  30,  a:)  *  times 
per  minute.    In  what  time  does  one  of  the  spokes  turn  through  a 
right  angle  ? 

18.  A  screw  required  10^  complete  turns  before  it  was  firm  in  the 
wood.    The  depth  of  the  hole  was  found  to  be  f  in.    How  far  did 
a  turn  of  a  straight  angle  drive  it  ? 

19.  What  is  the  complement  of  24°  17'?  of  79°  11'?  of  46°  34' 10"? 

20.  Find  the  sum  of  the  following  angles  :  26°  47'  3",  44°  22'  32", 
68°  51'  48",  39°  58'  37". 

21.  Find  the  supplements  of  the  following  angles :  163°  17',  48°  34', 
94°  52'  21". 

22.  The  supplement  of  an  angle  is  8  times  its  complement.    Find 
the  value  of  the  angle  in  degrees,  minutes,  and  seconds. 

23.  Reduce  to  degrees,  minutes,  and  seconds  •£•§  of  a  straight 
angle. 

*  Each  of  the  values  given  in  the  parenthesis  is  to  be  substituted  in 
place  of  the  value  given  before  the  parenthesis,  making  a  new  exercise. 


EXERCISES 


25 


24.  The  accompanying  figure  shows  the  card  of  the  mariner's 
compass.  Count  the  number  of  equal  parts  into  which  the  "points 
of  the  compass  "  divide  the  round  angle.  How  many  degrees  in  each 
of  these  parts?  Draw  a  compass  card. 


25.  Determine  by  how   many  degrees  the  following  directions 
differ : 

N.  and  E.  W.  and  N.E. 

N.  and  W.  N.N.E.  and  E.S.E. 

N.  and  S.  N.W.  and  S.E. 

S.  and  S.W.  N.W.  and  S.W. 

N.  and  E.S.E.  S.S.E.  and  W.N.W. 

W.S.W.  and  S.E.  S.W.  by  S.  and  S.W.  by  W. 

S.  and  N.W.  N.W.  by  W.  and  S.  by  E. 

26.  Determine  the  course  of  a  ship,  that  is,  the  point  of  the  com- 
pass toward  which  a  ship  is  sailing,  in  each  of  the  following  cases : 

N.  45°  W.  S.  45°  E.  N.  67^°  W.  S.  22^°  E. 

S.  111°  W.  N.  90°  E.  S.  67£°  E.  S.  56^°  W. 

NOTE.    N.  45°  E.  means  that  the  ship  is  sailing  northeast.    It  signifies 
a  direction  45°  east  of  north. 


26      PLANE  GEOMETRY— PRELIMINARY  COURSE 


EXERCISES 
REVIEW  AND  EXTENSION 

1.  From  a  point  0  as  a  common  origin  draw  six  rays  indicated 
in  order  by  the  letters  A,  B,  C,  D,  E,  F.    Express  Z  A  OD  and  Z  EOF 
each  as  the  sum  of  three  angles ;  Z  A  OC 

and  /.BOD  each  as  the  difference  of 
two  angles;  ZAOB  +  /.BOD— /.COD 
as  a  single  angle. 

2.  If  ZAOB  =  ZBOC,  what  name  is 
given  to  OB  ? 

3.  In  the  above  figure, 

if  ZAOB  =  /.COD,  show  that  ZAOC  =  ZBOD; 
if  ZAOC  =  /.BOD,  show  that  ZAOB  =  /.COD; 
if  /.A OB  =  Z  COD,  while  /.A OD  =  3  ZA  OB  and 

ZBOE  =  3Z  COD,  show  that  ZAOD  =  ZBOE; 
it  ZAOB  =  $ ZAOC,  and  Z  COD  =  ±Z  COE, 
while  ZAOC  =  Z  COE,  show  that  ZAOB  =  Z  COD. 

Express  in  words  the  principles  underlying  these  relations. 

4.  On  one  side  of  a  straight  line,  and  having  a  common  vertex, 
are  situated  (a)  three  (four,  five)  equal  angles;  (b)  four  angles,  of 
which  each  after  the  first  is  twice  as  large  as  the  preceding  one. 
How  many  degrees  in  each  angle  ? 

Let  x  =  number  of  degrees  in  the  first  angle. 

5.  There  are  four  angles  about  a  point,  of  which  each  after  the 
first  is  three  times  as  large  as  the  preceding  angle.    How  many  degrees 
in  each  angle  ? 

6.  What  is  the  result  in  Ex.  5,  if  the  second  angle  is  three  times 
the  first,  and  the  third  and  fourth  are  each  equal  to  the  sum  of  the 
two  preceding  angles? 

7.  One  of  two  supplementary  angles  is  2  (3,  4)  times  as  large  as 
the  other.    How  many  degrees  in  each  angle  ? 

8.  There  are  five  angles  about  a  point.    Four  of  them  have  the 
following  magnitudes  :  74°,  101°,  59°,  94°.    How  many  degrees  in  the 
fifth  angle  ? 


EXERCISES 


27 


9.  Given  A  a  =  40°,  Z  ft  =  30°,  Z  c  =  20°,  Z  rf  =  10°.    Draw  angles 
equal  to  Z  a  +  Z  b,  Z  6  +  Z  d,  Z  &  -  Z  e,  Z  6  +  Z  a  -  Z  d. 

10.  Given  Z  a  =  150°,  Z  b  -  60°.  Draw  angles  equal  to  ~^—  -  ; 


ro> 


11.  How  many  degrees  does  the  minute  hand  of  a  clock  traverse 
in  1  hr.?  in  \  (J,  j)  hr.?  in  30  (20,  10,  25)  min.  of  time?  in  5  (12, 
24,  36)  min.  of  time  ? 

12.  In  what  time  does  the  minute  hand  of  a  clock  describe  an 
angle  of  45°?  90°?  180°?  270°? 

13.  Change  to  the  lowest  indicated  denominations  20°  24';  30°  30'; 
179°  59'  60". 

14.  Draw  an  angle  of  50°  and  bisect  it 
with  the  aid  of  the  protractor. 

15.  In  the  figure  in  and  n  are  the  bisectors 
of  the  two  suppleinjsntary-adjacent  angles. 
Prove  that  m  -L  n. 

16.  If  an  angle  is  bisected,  prove  that  the  supplements  of  the  two 
equal  angles  are  equal. 

17.  What  kind  of  angle  is  formed  by  the 
bisectors  of  two  vertical  angles? 

18.  In  the  figure  m  JL  n,  and  /.a  —  /.a'. 
Show  that  Z  b  =  Z.  c. 

19.  The  sum  of  an  acute  angle  and  an  ob- 
tuse angle  cannot  exceed  how  many  degrees  ? 

20.  If  /.a  is  greater  than  Zb,  show  that  the  supplement  of  /.a 
must  be  less  than  the  supplement  of  Z.I. 

21.  How  many  degrees  in  an  angle  which  is  12°  less  than  its  sup- 
plement ?   18°  greater  than  its  complement? 

22.  Find  Z  A    and  Z  B,  if  l  (Z  A  +  Z  5)  =  48°   16'   20",   while 
\(£A  -Z  5)  =  22°  52'  17". 

23.  Find  the  value  of  one  half  the  supplement  of  65°  11'  31". 

24.  How  many  complete  revolutions  in  an  angular  magnitude  of 
7832°  ?   How  many  degrees  between  the  initial  and  final  positions  of 
the  revolving  ray  ? 


28      PLANE  GEOMETRY— PRELIMINARY  COURSE 


TRIANGLES 

45.  Experience  teaches  us  that  two  straight  lines  do  not  in- 
close a  space.  If,  however,  the  sides  of  an  angle  are  crossed 
by  a  straight  line,  we  obtain  a  closed 
space. 

A  triangle  is  a  figure  bounded  by  three 
straight  lines.    The  points  of  intersec- 
tion of  the  lines  are  called  the  vertices, 
and  the  segments  joining  the  vertices  are  called  the  sides  of 
the  triangle. 


The  vertices  are  denoted  by  capital  letters  and  the  sides  by  small  let- 
ters. Sometimes  it  is  convenient  to  use  corresponding  letters,  that  is,  the 
side  opposite  Z  A  is  denoted  by  the  corresponding  small  letter  a. 

The  sum  of  the  three  sides  of  a  triangle  is  called  the 
perimeter. 

46.  The  Angles  of  a  Triangle.  Exterior  Angle.  The  following 
exercises  are  intended  to  develop  a  clearer  insight  into  the 
mutual  relations  of  the  angles  of  a  triangle. 

EXERCISES 

1.  Point  out  triangles  in  the  exterior  of  a  house;   in  the 
street ;  on  the  surfaces  of  certain  solids  (e.g.  pyramids). 

2.  Draw  five  triangles  of  different  shapes,  and  measure  the 
sides  of  each.    Tabulate  your  results.    Can  any  three  segments 
be  used  for  the  sides  of  a  triangle  ? 

3.  Measure  the  angles  of  the  triangles  and  tabulate.   You 
will  find  that  the  sum  of  the  three  angles  in  each  case  is  equal 


TKIANGLES 


29 


to  180°.  It  will  be  proved  later  that  this  is  true  of  all  triangles ; 
that  is,  that  the  sum  of  the  angles  of  any  triangle  equals  a 
straight  angle. 

An  exterior  angle  of  a  triangle  is  the  angle  formed  by  one 
side    of    the    triangle    and 
the  prolongation  of  another 
side. 

4.  In  the  figure  /.x  is  an 
exterior    angle,    while    the 

angles    a,  b,   c,   are   interior        /  \y 

angles.    Draw  the  figure. 

5.  How  many  exterior  angles  can  a  triangle  have  ?   Compare 
/.x  and  Z  y.  How  many  pairs  of  vertical  angles  in  the  figure  ? 

6.  Draw  these  triangles,  using  the  given  measurements : 


AB 

BC 

CA 

Z.A 

z/?  - 

zc 

4  in. 

5  in. 

30° 

7  cm. 

6  cm. 

40° 

40  mm. 

70  mm. 

80° 

5  in. 

45° 

45° 

5  cm. 

90° 

60° 

4  in. 

60° 

60° 

8cm. 

77° 

46° 

2.8  cm. 

48° 

116° 

7.3  cm. 

12.1  cm. 

28° 

8  cm. 

72° 

72° 

6.7  cm. 

6.7cm. 

126° 

2  in. 

30° 

60° 

3|in. 

2f  in. 

32° 

30     PLANE  GEOMETRY— PRELIMINARY  COURSE 

7.  How  many  degrees  in  the  angles  of  a  triangle  if  the  ratio 
of  the  angles  is  1:2:3?  1:1:2?  2:3:5?  3:5:7? 

8.  How  many  right  angles  may  a  triangle  have  ?    How 
many  obtuse  angles  ? 

9.  How  many  acute  angles  must  every  triangle  have  ? 

10.  Can  a  triangle  be  drawn  whose  interior  angles  are  40°, 
50°,  100°  respectively  ?  80°,  70°,  40°  ?  90°,  40°,  50°  ? 

11.  Two  angles  of  a  triangle  are  together  equal  to  100°. 
What  is  the  value  of  the  third  angle  ? 

12.  Verify  by  measurement  the  fact  (to  be  proved  later)  that 
an  exterior  angle  of  a  triangle  is  equal  to  the  sum  of  the  two 
remote  interior  angles. 

CLASSIFICATION  OF  TRIANGLES  BASED  ON  ANGLES 

47.  A  triangle,  one  of  whose  angles  is  a  right  angle,  is  called  a 
right  triangle.  The  sides  inclosing  the  right  angle  are  called  the 
legs,  and  the  side  opposite  the  right  angle  is  called  the  hypotenuse. 


RIGHT  OBTUSE  ACUTE  EQUIANGULAR 

48.  A  triangle,  one  of  whose  angles  is  an  obtuse  angle,  is 
called  an  obtuse  triangle. 

49.  A  triangle,  each  of  whose  angles  is  an  acute  angle,  is 
called  an  acute  triangle. 

50.  Acute  triangles  and  obtuse  triangles  are  called  oblique 
triangles. 

51.  A  triangle  whose  three  angles  are  equal  is  called  an 
equiangular  triangle. 

The  word  "  triangle  "  is  often  replaced  by  the  symbol  A. 


TRIANGLES  31 

THE  ISOSCELES  TRIANGLE 

52.  On  the  sides  of  an  angle  lay  off  equal  segments  from 
the  vertex  and  join  their  extremities.  The  result  is  a  triangle 
having  two  equal  sides.  Such  a  triangle  is  said  to  be  isosceles. 
The  equal  sides  are  called  the  legs,  and  the 
other  side  is  the  base  of  the  triangle.  In  the 
figure  b  is  the  base. 

The  angle  opposite  the  base  of  an  isos- 
celes triangle  is  called  the  vertex  angle  of      . . 

the  triangle. 

The  angles  at  the  ends  of  the  base  of  an  isosceles  triangle 
are  called  the  base  angles  of  the  triangle. 


EXERCISES 

1.  Draw  an  isosceles  triangle  whose  vertex  angle  is  acute ;  right ; 
obtuse. 

2.  Draw  three  isosceles  triangles.  Measure  the  base  angles  in  each. 
What  do  you  observe  ? 

3.  Draw  two  intersecting  lines  and  form  isosceles  triangles  lying 
on  opposite  sides  of  the  intersection  point. 

4.  A  plane  divides  space  into  two  regions  which  lie  on  opposite 
sides  of  the  plane.    A  line  in  a  plane  divides  the  plane  into  two 
regions  such  that  the  segment  joining  points  on  opposite  sides  of  the 
line  crosses  that  line.   Into  how  many  regions  do  the  bounding  lines 
(produced)  of  a  triangle  divide  the  plane  ? 

5.  Give  illustrations  of  isosceles  triangles  (for  example,  gable  of  a 
house,  faces  of  a  pyramid). 

6.  Show  by  a  drawing  that  if  an  unlimited  straight  line  intersects 
one  side  of  a  triangle  at  any  point  except  an  extremity,  it  must  pass 
through  the  interior  of  the  triangle  and  cut  one  of  the  other  sides. 

7.  A  line-segment  is  drawn  from  one  vertex  of  a  triangle  to  the 
opposite  side,  dividing  the  triangle  into  two  triangles.    If  one  of 
these  triangles  is  acute,  what  is  the  other  ?  If  one  is  right,  what  is 
the  other? 


32     PLANE  GEOMETRY— PRELIMINARY  COURSE 

8.  Through  one  vertex  of  a  triangle  draw  a  straight  line  falling 
without  the  triangle.    What  is  the  sum  of  the  three  angles  formed 
at  that  vertex  ? 

9.  Through  the  three  vertices  of  a  triangle  draw  lines  falling 
without  the  triangle,  so  that  a  new  tri- 
angle is  formed.    This  triangle  is  said  to 

be  circumscribed  about  the  given  triangle. 
The  original  triangle  is  said  to  be  inscribed 
in  the  new  triangle. 

10.  Lay  off  on  a  straight  line  lengths 

equal  to  the  perimeters  of  the  inscribed  and  of  the  circumscribed 
triangle.  How  do  they  compare  ? 

11.  If  in  the  adjoining  figure  Z  a  =  Z  a', 
prove  that  Z  ft  =  Z  ft'. 

12.  Produce  two  sides  of  a  triangle  be- 
yond their  point  of  intersection,  making  / 

the  extensions  equal  respectively  to  the  sides.  Join  the  extremities 
of  the  extensions.  Show  with  tracing  paper  that  the  new  triangle 
is  a  repetition  of  the  first. 

13.  If  in  the  adjoining  figure  Za  +Zft  is  less 
than  2  rt.  A,  prove  that  Z  c  +  Z d  is  greater  than 
2  rt.  A.    On  which  side  of  the  transverse  line 
will  the  other  two  lines  meet  when  produced  ? 

14.  A  straight  railroad  track  extending  di- 
rectly away  in  the  line  of  vision  appears  to  vanish  in  the  distance 
at  a  single  point.    What  geometrical  figures  do  the  rails  and  the 
ties  appear  to  form  ?  ^ 

15.  If  the  triangles  ABB'  and  ACC'  in  the 
adjoining  figure  are  isosceles,  what  is  the  rela- 
tion between  EC  and  B'C'1  Give  a  reason  for  B/_ \B' 

your  answer.  / \^, 

/  ~\ 

16.  Draw  five  rays  from  a  common  origin.     '  x 

On  each  lay  off  three  given  segments,  a,  ft,  c,  in  succession,  begin- 
ning at  the  origin.  Connect  the  corresponding  extremities.  Of 
what  does  the  figure  remind  you?  How  many  isosceles  triangles 
in  the  figure? 


PRELIMINARY  COURSE  33 

POLYGONS 

53.  A  plane  figure  may  be  bounded  by  three  sides,  four  sides, 
five  sides,  etc.   The  general  name  for  such  a  figure  is  "  polygon." 

A  polygon  is  a  plane  figure  bounded  by  straight  lines.  The 
sides,  the  vertices,  the  interior  and  the  exterior  angles,  and 
the  perimeter  of  a  polygon  are  defined  as  in  the  case  of  a 
triangle.  Two  angles  at  the  extremities  of  the  same  side  are 
said  to  include  that  side. 

54.  Polygons  classified  as  to  the  Number  of  Sides.    Polygons 
are  named  according  to  the  number  of  their  sides.    A  quadri- 
lateral has  four  sides ;  a  pentagon,  five  sides ;  a  hexagon,  six 
sides ;    an    octagon,    eight    sides ;    a 

decagon,    ten    sides ;     a    dodecagon, 
twelve  sides,  etc. 

55.  A  diagonal  of  a  polygon  is  a 
segment  joining  two  vertices  that  are 
not  consecutive.    In  the  figure  AC, 
AD,  AE,  are  diagonals. 

Unless  the  text  contains  a  remark  to  the  contrary,  we  shall  use  only 
polygons  each  of  whose  interior  angles  is  less  than  a  straight  angle.  Such 
polygons  are  called  convex  polygons. 

56.  Angles;  Diagonals.    The  relations  existing  between  the 
sides,  the  angles,  and  the  diagonals  of  a  polygon  may  be  readily 
inferred  from  such  exercises  as  the  following : 

EXERCISES 

1.  Draw  a  quadrilateral  and  produce  the  sides  beyond  the 
vertices.    How  many  exterior  angles  are  formed  ?    How  many 
pairs  of  vertical  angles  ? 

2.  Answer  the  same  questions  for  polygons  of  5,  6,  7,  8,  9, 
10,  n  sides. 

3.  How  many  diagonals  can  be  drawn  from  one  vertex  of  a 
polygon  of  4  (5,  6,  7,  8,  9, 10,  n)  sides  ?    Tabulate  your  answers. 


34     PLANE  GEOMETRY— PRELIMINARY  COURSE 


4.  In  the  previous  exercise  how  many  triangles  are  formed 
by  these  diagonals  ? 

5.  How  many  different  diagonals  can  be  drawn  in  a  quadri- 
lateral ?  in  a  pentagon  ?  in  a  hexagon  ?  Tabulate  your  answers. 

6.  How  many  pairs  of  vertical  angles  do  these  diagonals 
form  within  the  polygons  ? 

7.  Join  any  point  within  a  polygon  to  all  the  vertices.    How 
many  triangles  are  formed  ? 

8.  Join  any  point,  not  a  vertex,  in  the  perimeter  of  a  polygon 
to  all  the  vertices.    How  many  triangles  are  formed  ? 

57.  Regular  Polygons.  Since  60  is  contained  exactly  in  360, 
a  set  of  six  angles,  each  equal  to  60°,  can  be  drawn  around  a 
point.  If  on  the  sides  of  these  equal  angles  equal  segments  are 
laid  off  from  the  common  vertex,  and  the  points  of  division 
are  joined  in  order,  a  polygon  is  formed.  In  the  same  manner 
angles  of  90°,  72°,  45°,  36°,  may  be  used.  It  will  be  proved 
later  that  the  sides  and  the  angles  of  such  polygons  are  equal ; 
for  that  reason  they  are  called  regular  polygons. 


90 


90° 


90° 


EXERCISES 

1.  With  the  aid  of  the  protractor  draw  regular  polygons  of  4,  5, 
6,  8,  10  sides. 

2.  Test  your  drawings  by  measuring  their  sides  and  angles. 

3.  How  many  degrees  in  each  of  the  interior  angles  of  these 
polygons?  in  each  of  the  exterior  angles?   Tabulate  your  answers. 

4.  In  which  of  these  polygons  do  the  rays  form  straight  lines?  Why? 

5.  Give  concrete  illustrations  of  regular  polygons. 


REGULAR  POLYGONS 


35 


6.  With  the  aid  of  ruler,  protractor,  and  compasses  copy  and  ex- 
tend the  following  patterns,  which  are  frequently  used  in  tiling  or 
parquet  flooring.  Very  artistic  effects  may  be  secured  by  coloring 
the  parts  of  the  figures  in  accordance  with  some  definite  plan. 


7.  Draw  the  following  patterns.    In  the  first  figure  a  regular  hex- 
agon is  surrounded  by  rectangles  and  equilateral  triangles. 


8.  In  the  following  patterns  irregular  hexagons  intervene  between 
the  regular  polygons.    Draw  the  figures. 


NOTE.  Interesting  applications  of  geometry  are  furnished  by  the  field 
of  ornamentation  and  design.  It  has  been  found  that  the  earliest  deco- 
rative patterns  of  nearly  all  primitive  races  are  of  a  geometric  char- 
acter. These  patterns  usually  grew  out  of  such  arts  as  weaving,  the 
making  of  vases,  or  the  laying  of  tile  floors.  The  American  Indians 
skillfully  introduced  simple  designs  in  their  basketry  and  pottery,  and 
some  Mexican  rock  temples  show  profuse  geometric  embellishment. 


36      PLANE  GEOMETRY— PRELIMINARY  COURSE 


n 


THE  CIRCLE 

58.  Preliminary  Definitions.    A   regular  polygon  of  n  sides 
may  be  obtained  by  drawing  from  a  point  n  rays  forming  n 
equal  angles,  and  laying  off  on  these 

rays  equal  segments.  The  points  of 
division  (vertices)  A,  B,  C,  .  .  .  could 
obviously  have  been  obtained  more 
rapidly  by  drawing  a  circle.  The 
common  origin  is  the  center,  and 
each  of  the  segments  a  radius  of 
the  circle. 

59.  A  circle  is  a  closed  curved  line  in  a  plane,  all  points  of 
which  are  equally  distant  from  a  fixed  point  in  the  plane, 
called  the  center. 

Many  writers  use  the  term  "circumference"  in  the 
above  sense,  a  circle  being  defined  as  a  plane  figure 
bounded  by  a  circumference.  There  is  good  authority 
for  our  definition,  which  has  the  merit  of  agreeing  with 
common  usage  and  the  language  of  higher  mathematics. 

The  circle  plays  a  very  important  part  in  geometry.  It  is  the  only 
curved  line  we  shall  have  occasion  to  study  in  this  text.  Its  use  in  many 
of  the  most  fundamental  constructions  will  prove  its  importance.  Hun- 
dreds of  concrete  illustrations  of  the  circle  might  be  given  here  (e.g. 
ring,  coins,  dial  of  clock,  wheel,  protractor).  We  call  attention  espe- 
cially to  the  "  round"  bodies—  the  sphere,  the  cylinder,  the  cone  —  from 
which  circular  sections  may  be  obtained  (e.g.  sections  of  a  ball,  tree, 
megaphone). 

A  line-segment  connecting  the  center  and  any  point  of  the 
circle  is  called  a  radius.  Any  line-segment  passing  through  the 
center  and  having  its  extremities  in  the  circle  is  called  a 
diameter. 

From  these  definitions  it  follows  that : 

60.  All  radii  of  the  same  circle  are  equal.    All  diameters  of 
the  same  circle  are  equal. 

61.  Two  circles  having  equal  radii  or  equal  diameters  can  be 
made  to  coincide  and  are  equal. 


THE  CIRCLE  37 

62.  A  circle  divides  the  plane  into  an  interior  and  an  exterior 
region.    A  point  is  in  the  interior  or 

in  the  exterior  region,  according  as 
the  line  joining  it  to  the  center  is 
less  than  or  greater  than  the  radius. 
Any  interior  point  A  is  said  to  be 
within  the  circle,  while  any  exterior 
point  B  is  without  the  circle. 

63.  A  part  of  a  circle  (circumfer- 
ence) is  called  an  arc.    One  half  of  a 

circle  is  called  a  semicircle.    A  fourth  part  of  a  circle  is  called 
a  quadrant. 

To  draw  a  circle  we  use  a  special  instrument,  the  compasses. 
A  pencil  or  a  crayon  with  a  piece  of  string  attached  serves  the 
same  purpose. 

EXERCISES 

1.  How  would  you  locate  on  a  map  all  houses  that  are  1  mi. 
from  the  City  Hall  ? 

2.  What  is  the  meaning  of  the  statement,  "  The  earthquake 
was  felt  within  a  radius  of  100  mi.  "  ? 

3.  What  kind  of  path  does  the  extremity  of  a  clock  hand 
describe  ? 

4.  As  you  open  your  book  observe  the  path  described  by 
a  corner  of  the  cover. 

5.  How  does  a  gardener  make  a  circular  flower  bed  ? 

6.  Revolve  a  circular  piece  of  paper  about  one  of  its  diame- 
ters.   What  kind  of  solid  is  generated  ? 

7.  If  a  wheel  is  made  to  slide  along  the  axle  passing  through 
its  hub,  what  kind  of  solid  is  generated  ? 

8.  A  wheel  revolves  ten  times  per  second.  In  what  time  does 
a  point  on  its  rim  pass  through  a  quadrant  ? 

9.  In  what  time  does  the  extremity  of  the  minute  hand  of 
a  clock  describe  a  semicircle  ? 


38     PLANE  GEOMETRY— PRELIMINARY  COURSE 

10.  A  string  fitting  exactly  around  a  bicycle  wheel  was  found 
to  be  1\  ft.  long.    If  the  wheel  makes  three  revolutions  per 
second,  what  is  the  distance  traversed  in  one  minute  ? 

POINTS  AND  CIRCLES 

11.  Show  that  a  point  may  be  in  any  one  of  three  different 
positions  with  reference  to  a  circle. 

12.  Given  a  point  P.    Find  all  the  points  that  are  2  cm. 
from  P. 

13.  Given  a  circle  and  a  point  P.    Find  the  points  on  the 
circle  that  are  3  cm.  from  P.    When  is  the  solution  impos- 
sible ? 

14.  Given  two  points  A  and  B.    Find  a  point  that  is  3  cm. 
from  A  and  4  cm.  from  B.    How  many  solutions  are  possible  ? 
When  is  the  solution  impossible  ? 

15.  Given  a  point.    Required  to  draw  a  circle  of  radius  2  cm. 
that  shall  pass  through  this  point.    How  many  solutions  are 
possible  ? 

64.  Lines  and  Circles.    The  different  relative  positions  of  a 
line  and  a  circle  are  illustrated  in  the  following  exercises. 

EXERCISES 

1.  Draw  a  circle  and  lay  your  pencil  on  the  paper  so  that 
it  passes  through  the  center.    Move  the  pencil  away  from  the 
center.    Observe  that  a  line  either  cuts  a  circle  in  two  points, 
or  touches  the  circle,  or  falls  entirely  without  the  circle. 

2.  Given  a  line  and  a  point  not  on  the  line.   Draw  three 
circles  having  the  given  point  as  center  and  illustrating  the 
fact  that  there  are  three  possible  positions    of  a   line  with 
reference  to  a  circle. 

3.  Given  a  line  I  and  a  point  P.   Required  to  find  on  I  the 
points  that  are  5  cm.  from  P.  When  do  you  obtain  two  solu- 
tions ?    one  solution  ?    no  solution  ? 


LINES  AND  CIRCLES 


39 


SUMMARY 

65.  A  secant  is  a  straight  line  of  unlimited  length  which 
intersects  a  circle. 

a  3 

66.  A  tangent  is  a  straight  line  of  un- 
limited length  which  has  one  and  only 
one  point  in  common  with  a  circle.    The 
point  is  called  the  point  of  contact  or 
point  of  tangency. 

67.  A  chord  is  a  line-segment  joining 
any  two  points  of  a  circle. 

In  the  figure  at  is  a  secant,  a2  is  a  tangent,  while  AB  is  a  chord. 

68.  A  secant  to  a  circle  intersects  the  circle  in  two  and  only 
two  points. 

69.  Two  Circles.   The  different  relative  positions  of  two  cir- 
cles, and  other  geometrical  properties,  are  developed  in  the 
following  exercises : 

EXERCISES 

1.  Cut  out  two  circular  pieces  of  paper,  making  their  radii 
3  cm.  and  5  cm.,  and  place  them  in  each  of  the  following 
positions :  (1)  one  falls  entirely  without  the  other ;  (2)  they 
touch  externally ;    (3)  they  intersect ;    (4)  they  touch  inter- 
nally ;  (5)  one  is  entirely  within  the  other ;   (6)  their  centers 
coincide. 

2.  Draw  two  unequal  circles  in  six  different  relative  posi- 
tions. 

3.  Draw   two   circles    whose 
centers  are  A  and  B.     Join  A 
and  B. 

4.  Draw  a  segment  AB  3  in. 
long.    On  it  indicate  any  two 

convenient  points  |  in.  apart.   Name  these  points  C,  D,  so  that 
the  order  of  the  points  is  A,  C,  D,  B. 


40     PLANE  GEOMETRY— PRELIMINARY  COURSE 

5.  About  A  as  a  center,  with  AD  as  a  radius,  and  also  about 
B  as  a  center,  with  EC  as  a  radius,  draw 
circles.    How  many  points  do  these  cir-    A_ 


—\ 1 B 

cles  have  in  common  ? 

6.  Repeat  Ex.  5,  using  A  and  B  as  centers,  and  taking  for 
radii  the  segments  AD  and  BD  respectively.  How  many  points 
are  common  to  the  two  circles  ? 

7.  Repeat  Ex.  5,  using  A  and  B  as  centers,  and  taking  for 
radii  A  C  and  BD  respectively.    How  many  points  are  common 
to  the  circles  ? 

8.  Will  the  results  of  Exs.  5,  6,  7,  be  the  same  if  the  dis- 
tances separating  A,  C,  D,  B,  are  changed,  and  only  the  order 
remains  the  same  ? 

9.  In  each  of  Exs.  5-7  state  the  relation  between  the  length 
of  the  segment  joining  the  centers  and  the  sum  of  the  radii  of 
the  two  circles. 

10.  If  the  lengths  of  the  segment  between  the  centers  and  of 
the  radii  of  two  circles  be  represented  by  I,  r,  r'  respectively, 
express  the  relation  between  I,  r,  and  r1  in  each  of  the  six  posi- 
tions of  two  circles  obtained  in  Ex.  2. 

11.  Draw  five  circles  with  the  same  center.    Give  concrete 
examples  of  such  circles. 

12.  Can  two  circles  which  have  the  same  center  touch  or 
intersect  ? 

70.  Center  Line ;  Center  Segment.    The  line  passing  through 
the  centers  of  two  circles  is  called  their  line  of  centers  or  center 
line,  and  the  Segment  of  the  center  line  joining  the  centers  is 
called  their  center  segment. 

71.  If  the  center  segment  of  two  circles  is  less  than  the  sum 
but  greater  than  the  difference  of  their  radii,  the  circles  intersect 
in  two  and  only  two  points. 

72.  Concentric  Circles.  Circles  having  the  same  center  are  said 
to  be  concentric. 


PRELIMINARY  COURSE  41 

EXERCISES 
1.  Make  a  careful  copy  of  the  two  following  figures : 


The  figures  are  made  up  of  semicircles  having  their  centers  in  the 
same  straight  line. 

2.  Draw  the  following  figure,  the  diameters  of  the  smaller  semi- 
circles being  half  as  large  as  those  of  the  greater  semicircles,  and  the 
centers  of  the  semicircles  being  in  the  same  straight  line. 


3.  On  a  straight  line  lay  off  in  succession  a  set  of  equal  segments. 
With  the  extremities  of  these  segments  as  centers,  and  with  a  radius 
equal  to  twice  one  of  the  segments,  draw  arcs  of  circles  as  shown  in 
the  figure.  The  smaller  arcs  are  constructed  from  the  same  centers. 
Draw  the  figure  and  complete  it. 


4.  Draw  a  circle  of  radius  3  cm.  Using  any  point  on  this  circle 
as  a  center,  and  with  the  same  radius,  draw  another  circle.  With 
each  of  the  two  intersection  points  on  the  first  circle  as  centers, 
keeping  the  same  radius,  draw  circles.  Repeat  the  process,  using 
only  points  on  the  first  circle  as  centers.  The  result  is  a  design 
frequently  used  in  ornamentation. 


42     PLANE  GEOMETRY— PRELIMINARY  COURSE 
5.  Draw  the  following  figures  : 


In  the  first  figure  the  semicircles  have  as  their  diameters  the  radii 
of  the  outer  circle.  The  second  figure  is  a  "  spiral,"  composed  of 
consecutive  semicircles,  each  new  radius  being  twice  the  previous 
radius  and  the  centers  lying  on  a  straight  line,  each  at  the  extremity 
of  the  arc  preceding. 

6.  Copy  and  complete  the  following  figures  : 


In  each  case  draw  two  mutually  perpendicular  straight  lines.  In 
the  first  figure  lay  off  on  the  four  rays  equal  consecutive  segments. 
Then  draw  four  sets  of  circles,  two  of  which  are  shown  in  the  diagram. 


CONSTRUCTIONS 


43 


73.  Construction  of  Triangles,  given  the  Sides.    If  the  extremi- 
ties of  a  segment  AB  are  used  as  the  centers  of  circles,  it  is 
obviously  possible  to  find 

two  radii,  AY  and  BX, 
such  that  the  two  circles 
intersect.  If  either  point 
of  intersection  is  joined 
to  A  and  B,  a  triangle  is 
formed. 

The  circles  will  inter- 
sect when  the  line  of 
centers  is  less  than  the  sum  of  the  radii;  that  is,  when 
AB<AC  +  BC.  We  infer  from  this  that  a  triangle  can  be 
constructed  with  three  given  lines  as  sides,  when  the  sum  of 
any  two  sides  is  greater  than  the  third  side. 

74.  Construction  I.    To  construct  a  triangle,  given  the  three 
sides. 


The  method  of  construction  is  shown  in  the  figure. 

75.  Scalene  Triangle.    A  triangle  in  which  no  two  sides  are 
equal  is  said  to  be  scalene. 

76.  Construction  II.    To  construct  an  isosceles  triangle,  given 
the  base  and  a  leg. 


I 


The  method  of  construction  is  shown  in  the  figure. 
Can  b  and  I  be  of  any  length  ? 

77.  The  vertex  C  in  the  figure  is  said  to  be  equidistant  from 
A  and  B. 


44     PLANE  GEOMETRY— PRELIMINARY  COURSE 


78,  Equilateral  Triangle.    A  triangle  of  which  all  three  sides 
are  equal  is  said  to  be  equilateral. 

79.  Construction  III.     To  construct  an  equilateral  triangle, 
given  a  side. 


AC  =  AB  =  EC,   Hence  the  three  sides  of  the  triangle  ABC 
are  equal. 

NOTE.   This  construction  is  the  first  proposition  of  Book  I  of  Euclid's 
Elements. 

EXERCISES 

1.  Construct  triangles,  when  possible,  from  the  following  data,  a,  b, 
and  c  representing  the  sides.  What  kind  of  triangle  results  in  each  case  ? 


a 

b 

c 

4  cm. 

5  cm. 

6  cm. 

3  in. 

4  in. 

5  in. 

4  in. 

5  in. 

9  in. 

7  cm. 

8  cm. 

8  cm. 

5  cm. 

5  cm. 

8  cm. 

3  in. 

3  in. 

3  in. 

1  in. 

2  in. 

4  in. 

2.  Construct  a  point  equidistant  from  two  given  points.    How 
many  solutions  are  possible  ? 

3.  Construct  six  different  isosceles  triangles  on  the  same  base. 
Join  their  vertices  in  succession.    What  do  you  observe  ? 


CONSTRUCTIONS  45 

4.  Construct  an  equilateral  triangle  ABC,  and  on  each  of  its 
sides  construct  an  equilateral  triangle.    Show  that  if  the  three  tri- 
angles surrounding  triangle  ABC  be  folded  over  on  the  sides  of  the 
original  triangle,  a  pyramid  can  be  made,  having  the  given  triangle 
ABC  as  its  base. 

5.  Draw  several  circles  so  that  each  passes  through  two  given  points. 

6.  Can  two  circles  intersect  in  three  points  ? 

7.  Draw  three  circles  such  that  each  intersects  the  other  two. 

8.  Draw  three  circles  such  that  each  passes  through  the  centers 
of  the  other  two. 

9.  Draw  four  circles  such  that  each  intersects  the  other  three. 

10.  Classify  triangles   according  to  their  angles;    according  to 
their  sides. 

11.  With  the  aid  of  ruler  and  compasses  draw  the  following  figures : 


In  the  first  figure  the  radii  of  all  the  arcs  are  equal,  the  centers 
being  respectively  the  vertices  of  the  square  and  the  mid-points  of 
the  sides.  The  second  figure  is  based  on  an  equilateral  triangle,  the 
centers  of  the  interior  arcs  being  the  mid-points  of  radii  drawn  to 
the  vertices  of  the  equilateral  triangle. 

12.  Lay  off  on  a  straight  line  AB  ten  successive  centimeter  divi- 
sions, and  mark  the  points  of  division  0, 1,  2,  3,  4,  •  •  • .  With  point  2 
as  a  center,  and  with  a  radius  of  2  cm.,  draw  a  semicircle  terminated 
by  points  0  and  4.  On  the  same  side  of  AB,  and  with  point  8  as  a 
center,  draw  a  semicircle  like  the  first.  With  point  5  as  a  center,  and 


46      PLANE  GEOMETRY— PRELIMINARY  COUKSE 

on  the  same  side  of  AB  as  before,  draw  a  semicircle  passing  through 
the  points  2  and  8.  Then,  using  the  points  1,  5,  and  9  as  successive 
centers,  and  with  a  radius  of  1  cm.,  draw  semicircles  on  the  opposite 
side  of  AB.  The  completed  figure  should  present  the  appearance  of 
a  closed  curve. 

13.  In  the  first  of  the  two  following  designs  locate  by  measure- 
ment the  centers  of  the  arcs  surrounding  the  equilateral  triangle. 
In  the  second  design  the  vertices  of  the  square  are  the  centers  of  the 
arcs.  Draw  the  figures. 


14.  With  the  aid  of  ruler  and  compasses  draw  the  following  figures  : 


In  each  of  these  figures  the  vertices  of  the  square  are  used  as  cen- 
ters for  four  of  the  arcs.  In  the  second  figure  the  radius  equals  one 
side  of  the  square. 

15.  Complete  the  circles  in  the  figures  for  Ex.  13. 


PRELIMINARY  COURSE 


47 


B 


CIRCLE  AND  ANGLE 
PRELIMINARY  DEFINITIONS 

80.  An  angle  formed  by  two  radii  is  called  a  central  angle. 

81.  A  central  angle  is  said  to  intercept  the  arc  cut  off  by 

its  sides. 

82.  A  chord  is  said  to  subtend  the  arc 
whose  extremities  it  joins. 

83.  Two  arcs  are  said  to  be  complements, 
supplements,     conjugates,     of    each    other, 
according  as  their  sum  is  a  quadrant,  a 
semicircle,  a  circle,  respectively. 

84.  Of  two  unequal  conjugate  arcs  of  a  circle,  the  less  is 
called  the  minor  arc  and  the  greater  the  major  arc.    A  minor 
arc  is  generally  called  simply  an  arc. 

85.  Relation  of  Central  Angles,  Arcs,  and  Chords.  Let  the  figure 
suggest  a  wheel  with  eight  spokes,  the 

angles  at  the  center  being  equal,  each  con- 
taining 45°.    Copy  the  figure  on  a  piece 
of  tracing  paper,  the  point  correspond- 
ing to  A  being  named  A',  etc.    Revolve 
the  duplicate  figure  around  the  center  0. 
The  circle  will  move  on  itself.  When  the 
point  A '  reaches  the  position  B,  point  B' 
will  fall  on  C,  etc.    Hence  the  arc  AB  is  seen  to  coincide  with 
arc  EC.    (Why  ?)    From  this  we  infer  that 
Equal  central  angles  in  a  circle  intercept 
equal  arcs.  Conversely,  equal  arcs  of  a  circle 
are  intercepted  by  equal  central  angles. 

86.  If   in  the  previous   figures  we  had 
drawn  the  chords  subtending  the  arcs  AB} 
BC,  etc.,  we  could  have  proved  at  the  same 

time  that  the  equality  of  the  arcs  involves  also  the  equality  of 


48     PLANE  GEOMETRY— PRELIMINARY  COURSE 

their  subtending  chords.  For  when  the  extremities  of  two  equal 
arcs  of  a  circle  coincide,  their  chords  also  coincide.  (Why  ?) 
Hence 

Equal  chords  of  a  circle  subtend  equal  arcs  ;  and  conversely, 
equal  arcs  are  subtended  by  equal  chords. 

87.  These  relations  of  arcs,  chords,  and  central  angles  evi- 
dently hold  also  for  equal  circles.    They  enable  us  to  construct 
an  angle  equal  to  a  given  angle ;  to  twice  a  given  angle,  etc. 

88.  Summary.    In  the  same  circle  or  in  equal  circles: 

1.  Equal  central  angles  intercept  equal  arcs. 

2.  Equal  arcs  are  intercepted  by  equal  central  angles. 

3.  Equal  chords  subtend  equal  arcs. 

4.  Equal  arcs  are  subtended  by  equal  chords. 

RESULTING  CONSTRUCTIONS 

89.  Construction  IV.    To  construct  an  angle  equal  to  a  given 
angle. 


Let  /.AOB  be  the  given  angle.  With  center  0  and  any 
radius  OA,  describe  a  circle.  With  0'  as  a  center,  and  using  a 
radius  O'A'  equal  to  OA,  describe  a  circle.  Make  chord  A'B' 
equal  to  chord  AB  by  drawing  an  arc  with  center  A'  and 
radius  AB.  Then  Z.A'0'B'  =  ZAOB.  (Why?) 

Is  it  necessary  to  draw  the  entire  circle  ? 


CONSTKUCTIONS 


49 


90.  Construction  V.   To    construct   a   triangle,  given   a   side 
and  the  two  adjoining  angles. 


The  diagram  explains  the  construction.    It  is  possible  when 
the  sum  of  Z.A  and  Z.B  is  less  than  180°. 

91.  Construction  VI.   To  construct  a  triangle,  given  two  sides 
and  the  included  angle. 

A B 

A C  C 


The  diagram  explains  the  construction.  It  is  possible  for  all 
values  of  A  B,  A  C,  and  Z.  A  less  than  180°. 

92.  Theory  of  the  Protractor.  Con- 
struct a  number  of  equal  angles  having 
a  common  vertex  0.  With  center  0 
and  arbitrary  radii  draw  concentric 
circles.  The  result  is  seen  in  the  figure. 
The  equal  angles  intercept  equal  arcs 
on  each  of  the  circles.  If,  now,  there 
had  been  360  equal  angles  about  0, 
that  is,  360  angle-degrees,  each  circle 
would  have  been  divided  into  360  equal 
arcs.  These  360  equal  circle  divisions 
are  also  called  degrees.  An  arc-degree 
is  -g-^  of  a  circle.  Thus  if  Z  AlOBl  contains  30  angle-degrees,  arc  ^1B1 
contains  30  arc-degrees.  This  explains  why  two  arcs,  e.g.  ^L1-B1  and  AZB2, 


50     PLANE  GEOMETRY  —  PRELIMINARY  COURSE 

may  contain  the  same  number  of  arc-degrees  without  being  equal.  It  also 
explains  why  circular  protractors  of  different  sizes  can  be  used  for  draw- 
ing an  angle  of  a  given  number  of  degrees.  In  general, 

93.  The  number  of  angle-degrees  in  the  central  angle  is  equal  to 
the  number  of  arc-degrees  in  the  intercepted  arc.    In  other  words, 

94.  A  central  angle  is  measured  by  its  intercepted  arc. 

95.  Regular  Polygons.    The  chords  of  the 

arcs  intercepted  by  equal  central  angles  of  a        D-  /? 

circle  are  equal.    Also  the  base  angles  of  the  ^/T^v 

isosceles  triangles  thus  formed  can  be  proved 

equal  by  folding.    Hence  the  doubles  of  these 

base  angles  are  equal ;  for  example,  /.ABC  = 

Z  BCD.   Whenever  the^  series  of  equal  chords 

forms  a  closed  perimeter,  a  regular  polygon 

results;  for  the  sides  and  the  angles  of  such 

a  polygon  are  equal.   This  will  always  happen 

when  the  first  central  angle  is  contained  an  integral  number  of  times  in 

a  round  angle. 

EXERCISES 
ANGLE  CONSTRUCTIONS 

(In  the  following  exercises  the  protractor  should  not  be  used.) 

1.  At  a  given  point  in  a  line  construct  an  angle  equal  to  a  given 
angle.    Show  that  this  may  be  done  in  four  ways. 

2.  Construct  an  angle  five  times  as  large  as  a  given  acute  angle. 

3.  Construct  an  angle  equal  to  the  difference  of  two  given  angles. 

4.  Construct  the  difference  of  an  interior  angle  of  a  regular  hexagon 
and  an  interior  angle  of  a  regular  pentagon.    Is  the  resulting  angle 
contained  exactly  in  a  round  angle  ? 

5.  Double  a  given  arc  of  a  circle. 

6.  Construct  the  difference  of  two  arcs  of  a  circle. 

7.  Construct  an  angle  equal  to  the  sum  of  the  angles  of  a  triangle. 
What  do  you  observe  ? 

8.  Construct  the  sum  of  the  angles  of.  a  quadrilateral. 

9.  Construct  the  sum  of  three  consecutive  angles  of  a  regular 
hexagon. 


EXERCISES 


51 


COMPUTATIONS 

10.  How  many  degrees  in  a  quadrant  ? 

11.  How  many  degrees  in  the  arcs  subtended  by  the  sides  of  a 
regular  polygon  of  4  sides  ?   5  sides  ?  6  sides  ?  10  sides  ?   15  sides  ? 
24  sides?   n  sides? 

12.  Through  how  many  degrees  does  a  point  on  the  equator  turn 
in  4  hr.  ?  6  hr.  ?  12  hr.  ?  18  hr.  ?  1  min.  ?  10  min.  ? 

13.  The  length  of  the  sun's  equator  is  about  2,722,000  mi.,  while 
that  of  the  earth's  equator  is  about  25,000  mi.    How  many  miles  in 
a  degree  of  each  of  these  equators  ? 

14.  The  sun's  time  of  rotation  is  about  25  days.  In  what  time  does 
a  sun  spot  near  the  equator  pass  through  72°  ?  How  many  miles  does 
it  traverse  during  that  time  ?   What  are  the  corresponding  answers 
for  a  point  on  the  earth's  equator  ? 

15.  If  the  earth's  orbit  around  the  sun  is  regarded  as  a  circle,  how 
many  degrees  does  the  earth  traverse  in  one  month?   in  one  day? 

16.  If  the  sun's  apparent  path  across  the  sky  each  day  is  regarded 
as  a  semicircle,  how  many  degrees  does  the  sun  traverse  in  1  hr.  on 
a  day  which  is  12  hrs.  long  ? 

REVIEW  AND  EXTENSION 

17.  Triangle  ABC  is  said  to  be  inscribed  in 
the  circle.   Inscribe  a  triangle  in  a  circle  and 
then  copy  the  figure. 

18.  How  may  a  segment  of  given  length  be 
made  a  chord  of  a  given  circle  ? 

19.  Lay  off  the  radius  of  a  circle  as  a  chord 
six  times  in  succession.  What  do  you  observe  ? 
Can  you  give  a  reason  for  this? 

20.  Construct  two  isosceles  triangles   on  opposite 
sides  of  a  common  base.   Join  their  vertices. 

21.  If  a  =  b  in  the  above  exercise,  show  that  each 
triangle  is  but  a  repetition  of  the  other.    Under  what 
circumstances  can  a  circle  be  circumscribed  about  the 
entire  figure  ? 


52     PLANE  GEOMETRY— -PRELIMINARY  COURSE 


22.  Inscribe  a  quadrilateral  in  a  circle.    Construct  the  sum  of  two 
opposite  angles  of  the  figure. 

23.  If  the  center  0  of  a  circle  is  joined  to  the 
extremities  of  a  chord  AB,  prove  that  an  isos- 
celes triangle  is  formed. 

24.  If  the  isosceles  triangle   OAB  in  Ex.  23 
revolves  about  0,  prove  that  A  and  B  move  on 
the  circle.    What  kind  of  figure  does  the  mid- 
point P  of  AB  describe? 

25.  In  a  semicircle  inscribe  a  triangle,  making  the  diameter  one 
side  of  the  triangle. 

26.  Prove,  by  folding,  that  any  diameter 
of  a  circle  bisects  the  circle. 

27.  Two  lines,  m  and  n,  intersect,  one  of 
the  angles  formed  being  Z  a.    At  any  point 
of    n    construct    Z  b  =  Z.  a.     Which    other 
angles  in  the  figure  are  equal  to  Z  a  and 
/.  b  ?    How  is  Z  x  related  to  Z  b  ? 

28.  Construct  a  quadrilateral  (see  accom- 
panying figure)  such  that 

Za  +  Z6  +  Zc  =  2rt.  A. 

29.  How  many  of  the  geometric  ideas 
developed  so  far  do  you  find  illustrated  con- 
cretely in  the  schoolroom  ?  /  ^ 

30.  Prepare  a  summary  of  the  most  important  definitions,  con- 
structions, and  propositions  in  Part  I. 


PART  II 


AXIAL  SYMMETRY 

96.  Preliminary  Exercises.  (1)  Fold  a  sheet  of  paper,  and 
before  opening  prick  a  hole  through  it.  Then  open.  There  will 
be  two  holes,  one  on  each  side  of  the  crease.  Name  these  A 
and  B,  and  name  the  line  of  the 
crease,  1.  Draw  the  segment  AB, 
and  name  the  point  where  it 
crosses  the  crease,  M. .  Show  that 

(1)  M  is  the  mid-point  of  AB-     A 

(2)  I  is  perpendicular  to  AB. 
The    line   of  folding   is    the 

perpendicular    bisector    of   the 
segment  AB. 

(2)  In  (1)  point  A  is  said  to  correspond  to  point  B.  This  is 
sometimes  indicated  by  the  abbreviation  A  \  B.  Extend  AB 
indefinitely.  Locate  three  pairs 
of  corresponding  points. 


M 


144 


A'  B'   C' 


In  the  figure,  A  \  A' 

B\B' 
C\  C' 

(Bead  "A  corresponds  to  .4'," 
etc.) 

(3)  Fold  a  piece  of  paper,  and  before  opening  prick  two  holes. 
Upon  opening  you  will  find  two  pairs  of 
holes.   Mark  these  A,A\  B,  B',  as  in  (2). 
Connect AzudA',  £and£'.  ThenAB\A'B' 
(read  "AB  corresponds  to  A'B"').   It  is 


B 


B' 


evident  that  to  every  point  on  AB  there 
corresponds  a  point  on  A  'B',  and  that  A  B  =  A  'B'.    Also  I  is  the 
perpendicular  bisector  of  A  A'  and  BB'. 

53 


54     PLANE  GEOMETRY— PRELIMINARY  COURSE 


(4)  Repeat  (3),  pricking  three  holes.  There  will  be  three  pairs 
of  corresponding  points.  Join  these  as  in  (3).  ABC  A'B'C'-,  that 
is,  to  every  point  of  A  ABC  there 
corresponds  a  point  of  AA'B'C'. 
A  ABC  may  be  made  to  coin- 
cide with  AA'B'C'.  In  the  same 
manner  we  could  obtain  a  poly- 
gon ABCDE--.  A'B'C'D'E'"-. 
The  two  figures  so  obtained  could 
be  made  to  coincide. 

SUMMARY 

97.  The  corresponding  points  in  the  preceding  figures  are 
often  called  symmetric  points.   The  line  of  folding  is  called  the 
axis  of  symmetry. 

98.  Two  points,  A  and  A\  are  said  to  be  symmetric  with 
respect  to  a  line  I,  called  their  axis  of  symmetry,  if  that  line  is 
the  perpendicular  bisector  of  the  segment  joining  the  two  points. 
In  general,  two  figures  are  symmetric  with  respect  to  a  line, 
called  their  axis  of  symmetry,  if  to  every  point  A  of  the  first 
there  corresponds  a  point  A '  of  the  second,  such  that  the  axis  is 
the  perpendicular  bisector  of  the  segment  joining  the  two  points. 

99.  Two  symmetric  (plane)  figures  can  be  made  to  coincide. 

Axial  symmetry  is  of  great  significance,  as  is  apparent  in  the  sym- 
metric structure  of  the  human  body,  of  animal  organisms,  of  leaves,  etc. 
Axial  symmetry  is  used  extensively  in  architecture,  designing,  pattern 
making,  etc. 

100.  Construction  of  Symmetric  Figures. 
It  is  evident  that  if  P  is  any  point  in  the 
axis,  PA  — PA'-,  that  is,  P  is  equidistant 
from  A  and  A '.   A  circle  of  radius  PA  and 
center  P  will  pass  through  A '.    This  circle 
must  intersect  the  axis  at  some  point  X. 

Then  XA  =  XA'.     These  relations  enable  us  to  construct  a 
point  A'  symmetric  to  a  given  point  A. 


x 


AXIAL  SYMMETRY 


55 


101 .  To  construct  A '  so  that  A'\A,  when  A  is  a  given  point  and 
I  is  the  given  axis.  From  any  point  P  in 
the  axis  as  a  center,  with  a  radius  PA, 
draw  an  arc  cutting  the  axis  at  X.  From 
X  as  a  center,  with  a  radius  XA ,  draw  an 
arc  intersecting  the  first  arc  at  A '.  A'  is 
the  symmetric  point  required.  (Why  ?) 

EXERCISES 


1.  Construct  a  segment  A'B*  so  that  A'Bf  \  AB. 

2.  Construct  a  triangle  A'B'C'  so  that  A'B'C'  \  ABC. 

3.  Construct  a  quadrilateral  A'B'C'D'  so  that  A'B'C'D'  \  A  BCD. 

4.  Place  a  plane  mirror  so  that  it  is  perpendicular  to  a  sheet  of 
paper.    Mark  a  point  on  the  paper.    Observe  that  the  point  and  its 
image  are  symmetrically  situated  with  respect  to  the  base  of  the 
mirror.    This  is  true  of  any  figure  drawn  on  the  paper.    Illustrate 
by  drawing  on  your  paper  a  square  ;  a  pentagon  ;  a  hexagon. 

5.  If  you  had  accidentally  closed  your  notebook  before  allowing 
the  ink  to  dry,  each  letter  would  have  made  a  symmetric  impression. 
How  is  the  same  principle  applied  in  printing  ? 

6.  Fold  a  sheet  of  paper  and  cut  out  any  figure  whatsoever,  be- 
ginning at  the  crease  and  returning  to  it.    A  symmetric  pattern  will 
be  obtained.   How  is  the  same  principle  applied  in  tailoring  ? 

7.  Study  the  capital  letters  of  the  alphabet.    How  many  do  not 
show  axial  symmetry  ? 

8.  What  geometrical  principle  is  illustrated  by  a  snow  crystal? 
(See  figures  below.) 

9.  The  figures  for  the  snow  crystals  show  that  more  than  one  axis 
of  symmetry  may  be  present.    How  many  are  there  in  each  case  ? 


56     PLANE  GEOMETEY— PRELIMINARY  COURSE 


102.  Conclusions.    The    following    statements    will   now  be 
readily  understood. 

1.  Any  point  in  the  axis  of  symmetry  of  two  points  A  and  B 
is  equidistant  from  them. 

2.  Any  point  equidistant  from  two  given  points  lies  in  their 
axis  of  symmetry. 

3.  Hence  the  axis  of  symmetry  of  two  given  points  contains 
all  the  points  equidistant  from  them. 

4.  Two  points  each  equidistant  from  the  extremities  of  a  seg- 
ment determine  its  perpendicular  bisector. 

The  equivalents  of  the  four  statements  pre- 
ceding will  be  considered  fully  in  Book  I. 

ft 


103.  Construction  of  the  Axis.    In  the 
figure,  P1?  P2,  P3,  etc.,  are  points  equi- 
distant from  A  and  B.  We  have  seen  that 
they  all  lie  in  the  perpendicular  bisector 
of  AB.  But  since  two  points  determine  a 
line,  two  of  these  equidistant  points  deter- 
mine the  perpendicular  bisector  of  AB. 

Hence,  to  construct  the  axis  of  sym- 
metry of  two  points,  A  and  5,  construct 
two  points  P  and  P',  each  equidistant 
from  A  and  B.  The  line  passing  through 
these  points  is  the  required  axis. 

104.  The  Isosceles  Triangle.    The  properties  of  the  axis  of 
symmetry  of  two  points  are  closely  con- 
nected with  the  isosceles  triangle.  In  the 

figure,  I  is  the  axis  of  A  and  B,  X  is  any 
point  of  the  axis,  M  is  the  mid-point 
of  AB.  The  truth  of  the  following  state- 
ments is  at  once  apparent. 

1.  X  is  the  vertex  of  an  isosceles  triangle  whose  base  is  AB. 

2.  Every  isosceles  triangle  has  an  axis  of  symmetry,  determined 
by  its  vertex  and  the  mid-point  of  its  base. 


FUNDAMENTAL  CONSTRUCTIONS 


5T 


3.  The  base   angles  (/-A  and  ZlJ3)  of  an  isosceles   triangle 
are  equal. 

Prove  by  folding. 

4.  The  per2>endicular  bisector  of  the  base  of  an  isosceles  tri- 
angle passes  through  the  vertex  and  bisects  the  vertex  angle. 

105.  The  Kite.  If  two  points  of  the  axis 
are  joined  to  the  symmetric  points,  we 
obtain  two  isosceles  triangles  having  a 
common  base.  Their  vertices  may  lie  on 
the  same  side  of  the  base  or  on  opposite 
sides.  The  result  in  the  latter  case  is  a 
kite-shaped  figure.  Since  these  figures 
will  serve  as  the  foundation  of  the  most 
fundamental  constructions,  we  shall  now 
give  a  table  of  their  most  important  prop- 
erties, using  the  accompanying  diagrams. 


Segments 

Angles 

a  =  a' 

Z  am  —  Z  a'  in' 

b  =b' 

Zbrn  =  Z  b'm' 

m  =  m' 

.-.  Z  ab   =  Z  a'b'. 

Why  ? 

Z  ad  =  Z  a'd 

Zbe    =  Zb'e 

\/ 


FUNDAMENTAL  CONSTRUCTIONS 

106.  To  bisect  a  given  segment. 
Construct  the  perpendicular  bisector 

of  AB. 

107.  At  a  given  point  of  a  straight  line 
to  erect  a  perpendicular  to  that  line. 

Take  the  given  point  P  as  a  center, 
and  with  any  convenient  radius   draw          x         p 
arcs  cutting  the  line  AB  at  X  and  Y.    Then  construct  Z  equi- 
distant  from  X  and  Y.    Join  P  and  Z.    PZ  is  the  required 
perpendicular.    (Why  ?) 


58     PLANE  GEOMETKY— PRELIMINAKY  COURSE 


X 


108.  By  the  previous  construction  it  is  pos- 
sible to  make  a  right  triangle.  Such  a  triangle 
may  be  made  of  cardboard  or  wood  and  used 
as  a  pattern  for  the  construction  of  perpen- 
diculars.  This  is  shown  in  the  figure.    R  is 

a  ruler  placed  along  the  given  line  and  T  is      - 
the  triangle.  ** 

109.  To  drop  a  perpendicular  to  a  given  straight  line  from  a 
given  external  point. 

FIRST  METHOD.    From  P  as  a  center, 
with  a  convenient  radius,  describe  an  arc 
cutting  AB  at  X  and  Y.    Then  construct 
Z  equidistant  from  A  and  Y.    PZ  is  the     A— 
required  perpendicular.    (Why  ?) 

SECOND  METHOD.    Use  a  right  triangle. 

110.  To  bisect  a  given  angle. 

From  P  as  a  center,  with  a  convenient 
radius,  describe  an  arc  cutting  the  sides  of 
the  '-angle  at  X  and  Y.  Construct  Z  equi- 
distant from  X  and  Y.  PZ  bisects  the  given 
angle.  (Why?) 

111.  To  bisect  a  given  arc. 

If  0  is  the  center  of  the  arc  AB,  bisect  the 
/.AOB.    OX  bisects  the  arc.    (Why?) 

\  *J        / 

NOTE.  The  five  constructions  of  §§  106-111  will  be  con- 
sidered again  in  Book  I.  At  this  point  they  are  intended  to 
give  practice  and  to  develop  greater  skill  in  using  the  geo- 
metric instruments,  viz.,  the  straightedge  and  the  compasses. 

EXERCISES 

1.  Show  by  repeated  bisection  that  a  segment  can  be  divided  into 
4,  8,  16,  2n  equal  parts. 

2.  An  angle  can  be  divided  into  2,  4,  8, 16,  2n  equal  parts.  Explain. 

3.  Construct  angles  of  90°,  45°,  22^°,  135°,  without  using  a  pro- 
tractor. 

4.  Construct  a  compass  card  without  using  a  protractor. 


EXERCISES 


59 


5.  Assuming  that  the  sum  of  the  angles  of  a  triangle  equals 
180°,  show  by  symmetry  that  each  angle  of  an  equilateral  triangle 
contains  60°. 

6.  Construct  an  angle  of  60°. 

7.  Construct   angles   of   30°,    15°,  75°, 
105°.    (75°  =  60°  +  15°.) 

8.  If  a  ray  of  light  yl  strikes  a  plane 
mirror  at  A,  it  is  reflected  in  such  a  manner 
that  Z-py-L  =  ^-Py^  j  that  is,  the  perpendicular 
p  to  m  at  A  bisects  Z.y^yy    Construct  the 
figure. 

9.  A  ray  of  light  is  reflected  by  a  mirror  m^  into  a  mirror  m^ 
Construct  the  path  of  the  ray. 

10.  The  figure  MNOQ  represents  a  billiard 
table.    Required  to  strike  the  ball  A  in  such  a 
manner  that  it  will  rebound  from  MN  and  hit 
the  ball  B.    Construct  the  path  of  A. 

Solution.  Construct  A'  \  A .  Draw  A'B,  cutting 
MN  at  P.   APE  is  the  path  of  the  ball  A. 

11.  We  can  now  construct  a  number  of  regular  polygons  without 
using  a  protractor.    The  first  figure  shows  how  a  square  maybe  con- 
structed, by  making  a  perpendicular  to  b,  etc.    The  second  figure 
shows  that  if  Z  0  =  60°,  &AOB  is  equilateral  (see  Ex.  5).    Hence 
AB  equals  OA,  the  radius  of  the  circle.    Copy  each  of  the  figures. 


12.  Construct  polygons  of  4,  8,  16  sides. 

13.  Construct  polygons  of  6,  12,  24  sides. 

14.  How  many  axes  of  symmetry  does  a  square  have?    a  regular 
pentagon  ?  a  regular  hexagon  ?  a  circle  ? 


60     PLANE  GEOMETRY— PRELIMINARY  COURSE 

15.  With  the  aid  of  ruler  and  compasses  draw  the  following 
figures : 


The  first  two  figures  are  based  on  the  fact  that  the  side  of  a  regu- 
lar hexagon  is  equal  to  the  radius  of  the  circumscribed  circle.  In  the 
third  figure  the  mid-points  of  the  sides  of  an  equilateral  triangle  are 
the  centers  of  the  arcs. 

16.  With  the  aid  of  ruler  and  compasses  draw  the  following- 
figures  : 


In  the  first  figure  the  vertices  of  the  six-pointed  star,  including 
the  points  of  intersection  of  the  sides  of  the  two  equilateral  triangles, 
are  the  centers  of  the  arcs.  The  second  figure  * 

arises  from  completing  the  arcs  of  the  third 
figure  of  the  preceding  exercise. 

17.  The  figure  shows  the  design  of  a  Gothic- 
window.  The  outer  arcs  A  C  and  A  B  are  con- 
structed with  B  and  C  as  centers.  The  small 
arcs  have  as  centers  the  points  B,  C,  D,  E,  F. 
The  center  0  is  obtained  by  drawing  arcs 

with  B  and  C  as  centers  and  with  a  radius  equal  to  BF.   Draw  the 
figure. 

(In  order  to  lessen  the  difficulty  of  this  exercise,  the  figure  should 
be  drawn  on  a  large  scale.) 


PRELIMINARY  COURSE 
CONGRUENCE 


61 


112.  Suppose  that  a. paper-cutting  machine  is  made  to  cut 
out  a  pattern  from  a  stack  of  paper.  The  pattern  will  then 
appear  as  many  times  as  there  are  sheets  of  paper.  Imagine 
all  these  figures  placed  along  a  line  I. 


The  figures  agree  in  all  respects,  since  they  are  but  repetitions 
of  one  figure.  This  could  at  any  time  be  verified  by  sliding  the 
figures  back  on  the  line  /  to  their  original  position  Fr  Each 
is  then  seen  to  be  a  duplicate  of  Fr  Such  figures  are  called 
congruent  (from  the  Latin  congruere,  "to  agree"). 

113.  Two  figures  which  may  be  made  to  coincide  in  all  their 
parts  are  said  to  be  congruent.    We  shall  use  the  symbol  =  for 
the  word  "  congruent."   Fl  =  F2  means  that  F^  is  congruent  to  F^. 
From  this  definition  it  follows  that  figures  congruent  to  the 
same  figure  are  congruent  to  each  other. 

114.  Homologous  Sides  and  Angles.    From  the  definition  it 
follows  that  if  two  figures  are  congruent,  each  side  and  angle 
of  one  has  a  corresponding  equal  side  and  angle  in  the  other. 

Thus  if  A  ABC  =  AA'B'C1, 

then  a  =  a 

and 


The  sides  and  angles  of  a  figure  are  called  its  principal  elements 
or  parts.  Any  two  corresponding  parts,  such  as  a  and  a',  are 
said  to  be  homologous. 


62     PLANE  GEOMETRY— PRELIMINARY  COURSE 
115.  Construction.   To  construct  a  polygon  congruent  to  a  given 


If  Pl  is  the  given  polygon,  construct 
A'B 


A'D'  =  AD, 
and      B'C1  =  EC. 


It  will  then  be  found  that  the  last  side  (C'D1)  and  the  two 
remaining  angles  (ZC'  and  ZZ>')  need  not  be  constructed,  as 
they  are  already  determined  or  fixed  by  the  preceding  elements 
of  the  figure.  The  new  polygon,  if  it  were  placed  on  the  given 
polygon,  would  coincide  with  it  exactly.  Hence  P2  =  Pr 

It  is  readily  seen  that  even  if  Pl  had  contained  a  larger  number  of 
sides,  we  should  have  arrived  at  the  same  conclusion  ;  namely,  that  the 
last  three  elements  of  a  polygon  are  determined  by  the  preceding  ones. 
Now  a  polygon  of  n  sides  has  n  angles,  or  2  n  parts.  Hence  two  polygons 
are  congruent  if  they  agree  in  2  n  —  3  consecutive  elements. 

116.   To  construct  a  triangle  congruent  to  a  given  triangle. 
Let  ABC   be  the   given   triangle,  in 
which 

a  =  6  cm.  c , 

b  =  7  cm. 

c  =  5  cm. 

FIRST  METHOD.  Draw  b1  =  b,  Z  A'  =  Z^1,Z  C'  =  Z  C.    Con- 
tinue the  sides  of  Z.A'  and  ZC'.    They 
will  intersect  at  a  point  B'. 

The  result  is  the  triangle  A'B'C'.  It 
can  be  shown  by  superposition  (that  is, 
by  placing  one  triangle  on  the  other) 
that  the  two  triangles  are  congruent. 

NOTE.  A  A'B'C'  may  be  constructed  congruent  to  A  ABC  with  the 
order  of  the  parts  the  same  or  the  reverse  of  that  in  A  A  BC  (compare  §  99). 


CONGRUENCE 


63 


c" 


SECOND  METHOD.  Draw  b'  =  b,  Z.A'  =  Z.A,  and  c'  =  c.  Con- 
nect B'}  the  extremity  of  c',  with  C', 
the  extremity  of  V.  A  definite  triangle 
A'B'C'  results.  The  congruence  of  the 
two  triangles  can  be  tested  by  super- 
position. 

THIRD  METHOD.  Draw  I'  =  b,  and 
with  centers  A'  and  C'  and  radii  c  and 
a  respectively,  describe  arcs.  These 
intersect  at  B'.  It  will  be  shown  later 
(§  142)  that  in  this  case  also 
AA'B'C'  =  AABC. 

117.  These  three  constructions  suggest  the  following  very 
important  laws  of  congruence  of  triangles : 

First  Law  of  Congruence.  If  two  triangles  have  a  side  and 
the  two  adjoining  angles  of  one  equal  respectively  to  a  side 
and  the  two  adjoining  angles  of  the  other,  the  triangles  are 
congruent,  (a.  s.  a.) 

Second  Law  of  Congruence.  If  two  triangles  have  two  sides 
and  the  included  angle  of  one  equal  respectively  to  two  sides 
and  the  included  angle  of  the  other,  the  triangles  are  congru- 
ent, (s.  a.  s.) 

Third  Law  of  Congruence.  If  two  triangles  have  the  three 
sides  of  one  equal  respectively  to  the  three  sides  of  the  other,  the 
triangles  are  congruent,  (s.  s.  s.) 

REMARK  1.  The  congruence  of  two  tri- 
angles cannot  be  made  to  depend  upon  angles 
only.  The  diagram  shows  that  triangles  may 
be  mutually  equiangular  without  being  con- 
gruent. 

REMARK  2.  The  frequent  use  of  these  laws  of  congruence  makes  ab- 
breviation desirable.  Thus  the  first  will  be  represented  by  the  abbrevia- 
tion "  a.  s.  a.,"  that  is,  the  law  of  one  side  and  the  two  adjoining  angles  ; 
the  second,  by  "s.  a.  s."  ;  the  third,  by  "s.  s.  s." 

REMARK  3.  Formal  proof  of  these  three  laws  is  given  among  the 
propositions  of  Book  I. 


64     PLANE  GEOMETRY— PRELIMINARY  COURSE 


EXERCISES 

1.  Construct  a  A  ABC  such  that  a  =  3  in.,  Z.B  =  45°,  Z  C  =  30°. 
Repeat  your  construction  and  test  the  congruence  of  the  triangles 
by  superposition  (a.  s.  a.). 

2.  Construct  A  A  EC  so  that  a  =  3  in.,  B  =  22^°,  c  =  5  in.    Repeat 
your  construction,  and  by  superposition  show  that  the  two  triangles 
coincide  (s.  a.  s.). 

3.  Construct  a  &ABC  such  that  a  =  4  cm.,  b  =  5  era.,  c  =  6  cm. 
Repeat  the  construction  and  test  with  tracing  paper  whether  or  not 
the  two  triangles  are  congruent  (s.  s.  s.). 

4.  Let  AB  represent  a  tower  100  ft. 
high.    Ascertain  by  measurement  from  a 
drawing  to  scale  how  large  an  angle  it 
subtends  at  the  eye  E  of  an  observer  if 
BE  =  200  ft.  ? 

5.  Construct  a  polygon  congruent  to  a  given  polygon  of  six  sides 
by  means  of  s.  s.  s. 

Suggestion.    Draw  diagonals  and  construct  congruent  triangles  in 
their  proper  positions. 

6.  Two  triangles,  in  order  to  be  congruent,  must 
have  three  parts  respectively  equal,  and  at  least  one 
side  must  be  used.    Explain. 

7.  What   is   meant  by   the   statement,  A  triangular   frame   is 
rigid  ?    Of  what  practical  value  is  this  fact  in  the 
construction  of  houses,  bridges,  truss-work? 

8.  Show  that  a  four-sided  frame  becomes  rigid 
if  a  diagonal  crosspiece  is  introduced.    How  does  a 
carpenter  use  this  fact  in  making  a  gate  ? 

9.  The  three  laws  of  congruence  may  be  replaced  by  the  state- 
ment :  A  triangle  is  determined  by  (1)  one  side  and  ^ 
two  angles  ;  (2)  two  sides  and  the  included  angle ; 

(3)  three  sides.    Explain. 

10.  Construct  a  quadrilateral  A  BCD,  (1)  if 
AB  =  2  in.,  EC  =  1  in.,  CD  =  f  in.,  DA  =  1%  in., 
^4C  =  2in.;  (2)  if  AB  =  5cm.,  BC=  6cm.,  CD  =  4 
AC  =8  cm. 


D 

cm.,  DA  =  5.6  cm., 


PRELIMINARY  COURSE 


65 


SURVEYING 

118.  Surveying  has  for  its  purpose  the  determination,  by 
processes  of  measurement,  of  the  relative  positions  of  points 
and  lines  upon  the  surface  of  the  earth.  A  careful  record  must 
be  kept  of  all  measurements  taken,  in  order  that  a  picture  or 
"  plat"  may  be  made  of  the  lines  or  areas  included  in  the  survey. 

119.  Lines  are  measured  with  chains,  tapes, 
or  rods. 

120.  The  angles  to  be  measured  directly 
are  either  horizontal  or  vertical. 

121.  Thus  if  A,  B,  C,  are  three  points  situated  in 
the  same  horizontal  plane  (for  example,  on  the  sur- 
face of  a  floor),  the  three  angles  J.,  B,  C  of  the 
triangle  ABC  are  horizontal  angles. 

Again,  if  EF  (see  figure  below)  represents  a  tower 
standing  on  a  horizontal  plane  DE,  the  angle  FDE, 
or  Z.  x,  is  vertical. 

122.  Angles  in  the  vertical  plane  are  either  angles  of  elevation 
or  angles  of  depression. 

A  person  standing  at  D  must  look  upward  in  order 
to  see  the  point  JP,  while  a  person  standing  at  .Fmust 
look  downward  in  order  to  see  the  point  D.  Hence, 
Zx  is  called  an  angle  of  elevation,  while  /.y  is  an 
angle  of  depression. 

123.  The  instrument  commonly  used  by  surveyors  for  meas- 
uring horizontal  and  vertical  angles  is  called 

a  transit.  It  consists  essentially  of  a  telescope 
free  to  move  in  a  vertical  circle  and  attached 
to  a  pointer  which  may  be  moved  over  a 
graduated  horizontal  circle.  The  circular  plate 
of  the  transit  can  be  made  level  by  suitable 
attachments. 

Let  the  center  of  the  plate  be  denoted  by  0.  Then 
if  two  points  A  and  B  are  in  the  same  plane  with 
0,  the  telescope  is  pointed  first  at  A  and  then  at  5,  and  the  differ- 
ence between  the  two  readings  on  the  graduated  circle  is  determined. 


D 


66     PLANE  GEOMETRY— PRELIMINARY  COURSE 


This  difference  is  the  value  of  the  horizontal  angle  AOB.  Vertical  angles 
are  measured  similarly  by  means  of  the 
vertical  circle. 

124.  Asimple  combination  instrument 
for  measuring  either  horizontal  or  verti- 
cal angles  is  shown  in  the  figure.  This  in- 
strument will  serve  as  an  inexpensive 
substitute  for  a  transit.  The  vertical  rod 
MM'  is  free  to  revolve  in  the  socket  at  Jlf, 
carrying  a  horizontal  pointer  which  in- 
dicates readings  on  the  horizontal  circle 
divided  into  degrees.  These  divisions  must 
be  marked.  The  pointer  at  M'  is  provided 
with  sights  and  is  free  to  move  in  a 
vertical  circle  around  M'.  By  sighting 
along  this  pointer,  vertical  angles  may  be 
measured. 

By  the  aid  of  some  such  instrument 
as  this,  and  of  a  50-ft.  tape,  a  few  rods 
and  some  stakes,  distances  and  angles  as  seen  from  some  given  point 
in  the  school  yard  should  be  measured. 


EXERCISES 

1.  In  the  diagram  A  and  B  are  points  separated  by  a  river.    How 
may  the  length  of  AB  be  determined  by  the 
first  law  of  congruence?  (a.  s.  a.)    (A,B,C, 
represent  trees  or  vertical  rods.    Measure  A  C 


B'CB  is  a  straight  line.  Measure  A  'B'.  Then  ' 
A'E'  =  AB.) 

2.  In  the  diagram  AB  represents  the 
distance  across  a  pond.  The  length  of  AB 
is  to  be  determined  by  the  second  law 
(s.a.  s.).  A  vertical  rod  is  placed  at  C.  Meas- 
ure AC  and  BC.  Make  A'C  =  AC,  and 
B'C  =  BC.  AC  A'  and  BCB'  are  straight 
lines.  Then  A'E  =  AB.  If  AC  =  300  ft., 
£(7  =  200  ft.,  ZC=-55°,  determine  AB  by 
measurement. 


SUEVEYING 


6T 


3.  Explain  what  measurements  are  necessary  in  order  to  deter- 
mine the  distance  of  an  anchored  ship  from  the  shore. 

Suggestion.    Use  the  method  of  a.  s.  a. 

4.  What  measurements  are  necessary  in  order  to  determine  the 
length  of  a  rectangular  building  whose  corners  only  are  accessible? 

Suggestion.    Use  the  method  of  s.  a.  s. 

5.  It  is  required  to  find  the  distance  between  two  ships  A  and  B 
anchored  off  the  shore. 

Solution.  Measure  a  convenient  length 
CD  along  the  shore.  With  the  instru- 
ment at  C,  and  then  at  D,  measure 
two  angles  at  each  point.  For  example, 
at  C,  measure  any  two  of  the  angles 
ACB,  ACD,  BCD.  Explain  how  AB  is  determined  by  these  five 
measurements. 

6.  Find  the  height  of  a  flagpole  by  measuring  a  convenient  dis- 
tance from  its  foot,  and  then  measuring  the  angle  of  elevation  of  the 
top  of  the  pole.    If  the  distance  is  150  ft.  and  the  angle  is  28°,  how 
high  is  the  pole  ?   Measure  by  drawing  to  scale. 

7.  Determine  the  height  of  your  high-school  building  by  a  draw- 
ing made  to  scale. 

8.  The  height  of  a  mountain  CD  may  be 
found  by  determining  its  angle  of  elevation 
from  two  different  points,  A  and  B,  situated 
in  the   same  vertical  plane  with   C.    Then, 
if  AB  is   known,    CD   can   be   constructed. 
Explain. 

9.  Explain  how  it  is  possible  to  find  from  the  ground  the  height 
of  a  kite,  a  balloon,  or  a  cloud. 

10.  Two  persons  600  ft.  apart,  situated  on  opposite  sides  of  a  bal- 
loon as  seen  from  the  ground,  observe  its  angles  of  elevation  to  be 
44°  and  55°.    What  is  the  height  of  the  balloon  ?   Draw  to  scale. 

11.  By  what  methods  can  the  distance  between  the  tops  of  two 
church  steeples  be  determined  ? 

12.  How    may   the    distance    between    two    mountain    tops    be 
determined? 


68     PLANE  GEOMETRY— PRELIMINARY  COURSE 


13.  Thales  of  Miletus  (600  B.C.)  is  said  to  have  invented  a  way  of 
finding  the  distance  of  a  ship  from  the  shore.    His  method  is  sup- 
posed to  have  been  based  on  the  law  of 

a.  s.  a.,  and  may  have  been  as  follows : 
Two  rods,  m  and  n,  are  hinged  together  at 
A.  One  arm,  m,  is  held  vertically,  while 
the  other,  n,  is  pointed  at  the  ship  S.  Then 
the  instrument  is  revolved  about  m  as  an 
axis  until  n  points  at  some  familiar  object 
£'  on  the  shore.  Then  A  ABS  =  A  A BS',  and  BS'  =  BS.  Explain. 

14.  The  bearing  of  a  place  A  from  another  place  B  is  the  angle 
of   deviation  of    the    line  BA    from  the  true   north-and-south  line 
through  B.    A  is  5  miles  west  of  B,  and  C  is  9  miles  south  of  A. 
Find,  by  measurement  with  the  protractor,  the  bearing  of  C  from  B  *, 
of  B  from  C.    (Use  a  scale  of  two  miles  to  the  inch.) 

15.  A  is  11  miles  from  B,  and  its  bearing  from  B  is  N.  17°  E. 
C  is  9  miles  from  A,  bearing  S.  43°  E.  Find,  by  construction  and 
measurement  with  ruler  and  protractor,  the  bearing  of  C  from  B; 
of  B  from  C;  and  the  distance  BC. 

Suggestion.  Draw  the  north-and-south  lines  through  the  given 
points.  These  will  be  at  right  angles  to  an  east-and-west  base  line 
at  the  bottom  of  the  figure. 

16.  A  surveyor  maps  a  field  by  taking  the  successive  bearings  and 
distances  from  one  corner  to  the  next.    Draw  a  map  from  the  follow- 
ing survey  notes,  using  a  scale  of  200  ft.  to  the  inch,  and  determine 
by  measurement  the  missing  data. 


Station  occupied 

Point  sighted  at 

Bearing 

Distance 

A 

B 

N.  34°  E. 

450  ft. 

B 

C 

N.  76°  E. 

200  ft. 

C 

D 

S.  12°  E. 

600  ft. 

D 

E 

S.  62°  W. 

225  ft. 

E 

A 

? 

? 

17.  The  bearing  of  Long  Branch,  New  Jersey,  from  Newark  is  S. 
18°  E.  and  its  distance  is  30  miles.  The  bearing  of  Trenton  from  Long 
Branch  is  S.  84°  W.  and  its  distance  is  41  miles.  Find  the  bearing  and 
the  distance  of  Newark  from.  Trenton.  (Use  any  convenient  scale.) 


BOOK  I 

RECTILINEAR  FIGURES 
A  SYSTEM  OF  PROPOSITIONS 

125.  In  the  preceding  pages  many  simple  geometric  prin- 
ciples were  presented  and  illustrated.    There  remain,  however, 
many  other  facts  of  geometry  which  are  not  so  apparent,  or 
which,  even  though  they  seem  quite  easy  of  comprehension, 
are  not  easily  proved. 

The  aim  of  demonstrative  geometry  is  to  point  out  methods  of 
discovery  and  of  proof,  as  well  as  a  logical  arrangement,  of  those 
geometric  principles  which  are  most  important  in  the  develop- 
ment of  the  subject,  and  which  have  the  widest  application 
in  other  fields  of  investigation.  Such  principles  are  called 
propositions,  and,  except  for  those  which  are  taken  for  granted 
at  the  outset,  are  established  by  means  of  formal  demonstrations. 

126.  A  complete  demonstration  demands  a  reason  for  every 
statement  which  is  included  in  it.    It  is  obvious  that  such  a 
justification  of  the  particular  steps  of  a  proof  is  greatly  facili- 
tated by  arranging  the  propositions  in  a  definite  progressive  order. 

127.  The  most  fundamental  propositions,  which  are  first  con- 
sidered, are  those  which  are  most  often  in  demand  as  the  work 
progresses ;  much  as  the  multiplication  table  is  needed  in  the 
study  of  arithmetic.    Other  propositions  are  shown  to  depend 
more  or  less  directly  upon  these,  and  thus  gradually  arises  a 
system  of  propositions,  each  of  which  depends  upon  some  or 
all  of  those  preceding,  in  the  order  of  consideration.    For  this 
reason  the  study  of  geometry  has  been  compared  to  the  build- 
ing of  a  house  or  other  structure ;  each  stage  of  the  work  is 
supported  by  what  has  already  been  completed. 

69 


70        PLANE  GEOMETRY— BOOK  I 

ARRANGEMENT  OF  A  DEMONSTRATION 

128.  The  formal  demonstration  of  a  theorem  in  geometry 
involves  three  steps.: 

1.  The  statement  of  what  is  given  (the  hypothesis). 

2.  The  statement  of  what  is  to  be  proved  (the  conclusion). 

3.  The  proof,  consisting  of  several  steps,  each  based  upon 
the  hypothesis,  or  upon  the  authority  of  definitions,  axioms, 
preliminary  theorems,  or  preceding  propositions. 

These  three  steps  are  indicated  by  the  words 

Given,  To  prove,  Proof. 

129.  In  order  to  avoid  a  mechanical  repetition  of  the  text, 
and  to  gain  the  ability  to  reason  independently  in  the  field  of 
geometry,  it  is  advisable  for  the  student  to  bear  in  mind  the 
following 

RULES 

1.  Read  the  proposition  carefully,  noting  the  meaning  of 
each  term  employed. 

2.  Draw  a  figure  which  represents  the  conditions  demanded 
by  the  proposition,  and  no  other  special  conditions. 

3.  State  the  hypothesis  and  the  conclusion  as  applied  to  that 
particular  figure. 

4.  Recall  the  geometric  processes  which  generally  lead  to 
this  conclusion,  and  examine  the  hypothesis  and  its  immediate 
consequences  in  order  to  discover  whether  such  processes  may 
be  applied  in  the  particular  instance. 

5.  Try  to  formulate  a  proof  before  examining  the  one  given 
in  the  text. 

6.  Write  out  the  proof,  arranging  the  steps  in  order  and 
numbering  the  main  divisions.    The  final  step  should  agree 
with,  the  conclusion.    After  each  step  give  the  authority  which 
supports  it.    Proper  authorities  are  the  hypothesis  and  any 
definitions,  axioms,    preliminary  theorems,  and  .propositions 


SYMBOLS  AND  ABBREVIATIONS  71 

previously  established.    Use  freely  such  symbols  and  abbrevi- 
ations as  are  suggested  below. 

7.  At  a  later  period  in  the  development  of  the  work  it  is 
often  sufficient,  and  even  preferable,  to  write  merely  a  brief 
summary  of  the  demonstration. 

8.  As  a  review  exercise,  practice  reciting  the  proof  from  a 
mental  figure  only. 

SYMBOLS  AND  ABBREVIATIONS 

+  plus,  or  added  to.  adj.  adjacent. 

—  minus,  or  diminished  by.  alt.  alternate. 

=  equal,  or  is  equal  to.  ax.  axiom. 

>  is  greater  than.  circum.  circumference. 

<  is  less  than.  comp.  complement,  or  complementary. 

=  is  congruent  to.  cons.  construction. 

.-.  therefore,  or  hence.  cor.  corollary. 

_L  perpendicular,  or  is  per-  corr.  corresponding, 

pendicular  to.  def .  definition. 

Js  perpendiculars.  ex.  exercise. 

II  parallel,  or  is  parallel  to.  ext.  exterior. 

Us  parallels.  horn.  homologous. 

-  is  similar  to,  or  similar.  hyp.  hypothesis. 

Z  angle.  int.  interior. 

A  angles.  isos.  isosceles. 

A  triangle.  prop,  proposition. 

A  triangles.  rect.  rectangle. 

O  parallelogram.  rt.  right. 

DI7  parallelograms.  sq.  square. 

O  circle.  st.  straight. 

©  circles.  supp.  supplement,  or  supplementary. 

a.  s.  a.,  having  a  side  and  the  two  adjoining  angles  of  one  equal  respec- 
tively to  a  side  and  the  two  adjoining  angles  of  the  other. 

s.  a.  s.,  having  two  sides  and  the  included  angle  of  one  equal  re- 
spectively to  two  sides  and  the  included  angle  of  the  other. 

s.  s.  s.,  having  three  sides  of  one  equal  respectively  to  three  sides  of 
the  other. 

rt.  A,  h.  1.,  being  right  triangles  and  having  the  hypotenuse  and  a 
leg  of  one  equal  respectively  to  the  hypotenuse  and  a  leg  of  the  other. 

rt.  A,  h.  a.,  being  right  triangles  and  having  the  hypotenuse  and  an 
adjoining  angle  of  one  equal  respectively  to  the  hypotenuse  and  an 
adjoining  angle  of  the  other. 


72  PLANE  GEOMETRY— BOOK  I 

AXIOMS.    PRELIMINARY  PROPOSITIONS 

130.  All   demonstration   finally    depends    on   taking   some 
truths  for  granted.    Such  truths  are  sometimes  called  axioms. 
They  are  assumed  to  be  so  clear  in  themselves  that  every  one 
will  accept  them  without  proof. 

Geometry  is  no  exception  to  this  general  rule.  It  is,  how- 
ever, essential  in  elementary  geometry  that  the  facts  taken 
for  granted  and  without  demonstration  should  be  of  a  simple 
nature,  so  simple,  in  fact,  as  to  be  obvious  if  merely  plainly 
stated.  The  geometrical  facts  which  are  assumed  in  Book  I 
will  now  be  tabulated  under  the  heading  Preliminary  Assump- 
tions and  Propositions,  and  it  will  be  observed  that  we  are 
thus  giving  a  recapitulation  of  some  of  the  main  results  of 
the  informal  discussion  of  the  Preliminary  Course. 

131.  Preliminary  Assumptions  and  Propositions. 

1.  Through  a  given  point  an  indefinite  number  of  straight 
lines  may  be  drawn  (§  11). 

2.  Two  straight  lines  can  intersect  in  but  one  point  (§  11). 

3.  One  and  only  one  straight  line  can  be  drawn  through  two 
given  points  (§  11). 

4.  Two  straight  lines  that  have  two  points  in  common  coin- 
cide throughout  and  form  but  one  line  (§  11). 

5.  Two  straight  lines  do  not  inclose  a  space  (§  11). 

6.  Two  line-segments  whose  extremities  can  be  made  to  coin- 
cide must  coincide  throughout  (§  13). 

7.  Two  line-segments  a  and  b  must  be  in  one  of  three  rela- 
tions to  each  other,  namely,  a  >  b,  a  =  b,  or  a  <  b  (§  13). 

8.  Line-segments  may  be  added  together.    Of  two  unequal 
line-segments  the  smaller  may  be  subtracted  from   the  larger. 
Line-segments  may  be  multiplied  by  a  given  number  (§  16). 

9.  All  round  angles  are  equal  (§  22). 

10.  All  straight  angles  are  equal  (§  24). 

11.  All  right  angles  are  equal  (§  28). 


AXIOMS  73 

12.  Two  angles  A  and  B  must  be  in  one  of  three  relations  to 
each  other:  Z.I  >  Z.B,  /.A  =  Z.B,  or  Z,4<Z£  (§  32). 

13.  At  a  given  point  in  a  given  line  only  one  perpendicular 
can  be  drawn  to  that  line  (in  the  same  plane)  (§  41). 

14.  The  complements  of  the  same  angle,  or  of  equal  angles, 
are  equal  (§  41). 

15.  The  supplements  of  the  same  angle,  or  of  equal  angles, 
are  equal  (§  41). 

16.  Vertical  angles  are  equal  (§  41). 

17.  If  two   adjacent  angles  have  their  exterior  sides  in  a 
straight  line,  they  are  supplementary  (§  41). 

18.  If  two  adjacent  angles  are  supplementary,  their  exterior 
sides  lie  in  a  straight  line  (§  41). 

19.  Radii  of  the  same  circle  are  equal  (§  60). 

20.  At  a  given  point  in  a  given  line  a  line  may  be  drawn 
'making  with  the  given  line  an  angle  equal  to  a  given  angle 
(§  89). 

21.  Figures  congruent  to  the  same  figure  are  congruent  to  each 
other  (§  113). 

132.  General  Axioms.    Further  fundamental  truths  involving 
magnitude  are  as  follows  : 

1.  Magnitudes  which  are  equal  to  the  same  magnitude,  or  to 
equal  'magnitudes,  are  equal  to  each  other. 

In  other  words,  A  magnitude  may  be  substituted  for  its  equal. 

2.  If  equals  are  added  to  equals,  the  sums  are  equal. 

3.  If  equals  are  subtracted  from  equals,  the  remainders  are 
equal. 

4.  If  equals  are  multiplied  by  equals,  the  products  are  equal. 

5.  If  equals  are  divided  by  equals,  the  quotients  are  equal. 

The  divisor  must  not  be  zero. 

6.  Like  powers  or  like  positive  roots  of  equals  are  equal. 

7 .  The  whole  of  any  magnitude  is  equal  to  the  sum  of  all  its 
parts. 

8.  The  whole  of  any  magnitude  is  greater  than  any  part  of  it. 


74  PLANE  GEOMETRY— BOOK  I 

CONGRUENCE 
PROPOSITION  I.    THEOREM 

FIRST    TRIANGLE   CONGRUENCE 

133.  If  two  triangles  have  a  side  and  the  two  adjoining 
angles  of  one  equal  respectively  to  a  side  and  the  two  adjoin- 
ing angles  of  the  other,  the  triangles  are  congruent,  (a.  s.  a.) 


Given  the  triangles  ABC  and  DEF,  in  which  AB  equals  DE, 
angle  A  equals  angle  D,  and  angle  B  equals  angle  E. 

To  prove  that  A  A  B  C  =  A  DEF. 

Proof.    1.  Place  A  ABC  on  A  DEF  so  that  AB  coincides  with 
DE,  and  C  and  F  fall  on  the  same  side  of  DE. 
2.  Then  AC  will  fall  along  DF, 


and  BC  will  fall  along  EF. 

(ZB=ZE,  by  hyp.) 

3.  /.  C  will  fall  upon  F.  §  11,  (2) 
(Two  straight  lines  can  intersect  in  but  one  point.) 

4.  /.  A  ABC  coincides  with  A  DEF, 

i.e.  A  ABC  =  A  DEF. 
(Definition  of  congruence,  §  113.) 

REMARK.  The  method  of  proof  adopted  is  superposition  (§  116),  that 
is,  "placing  on  "  ;  namely,  the  triangle  ABC  was  placed  on  the  triangle 
DEF.  The  so-called  axiom  of  superposition  has  been  tacitly  assumed. 
This  axiom  may  be  stated  thus  : 

Any  figure  may  be  moved  about  in  space  without  changing  either  its  size 
or  its  shape. 


CONGRUENCE  75 

PROPOSITION  II.    THEOREM 

SECOND    TRIANGLE   CONGRUENCE 

134.  If  two  triangles  have  two  sides  and  the  included  angle 
of  one  equal  respectively  to  two  sides  and  the  included  angle 
of  the  other,  the  triangles  are  congruent,  (s.  a.  s.) 


Given  the  triangles  ABC  and  DEF,  in  which  AB  equals  DE,  AC 
equals  DF,  and  angle  A  equals  angle  D. 

To  prove  that  A  ABC  =  A  DEF. 

Proof.    1.  Place  A  ABC  on  A  DEF  so  that  AB  coincides  with 
DE,  and  C  and  F  fall  on  the  same  side  of  DE. 

2.  Then  AC  will  fall  along  DF. 

(ZA  =  ZD,  by  hyp.) 

3.  Also       the  point  C  will  fall  on  the  point  F. 


4.  .'.  CB  will  coincide  with  FE.  §  13,  (4) 
(Their  extremities  being  the  same  points.) 

5.  .'.  A  ABC  coincides  with  A  DEF. 

.'.  AABC  =  ADEF. 
(Definition  of  congruence.) 

135.  Homologous  parts  (sides  or  angles)  of  congruent  tri- 
angles are  those  parts  which  are  opposite  parts  known  to  be 
equal.  Thus,  in  the  above  figures,  EC  is  homologous  to  EF, 
Z.B  to  Z.E,  and  Z  C  to  Z.F.  Obviously 

Homologous  parts  of  congruent  triangles  are  equal. 


76  PLANE  GEOMETRY— BOOK  I 

136.  The  congruence  of  two  triangles  affords  a  fundamental 
method  of  establishing  the  equality  of  two  given  lines  or  angles, 
as  outlined  in  the  following  scheme : 

In  order  to  prove  two  (  )  equal : 

\angles/ 

I.  Show  that  they  are  homologous  /  S1  ®s    \  Of  congruent  A. 

In  order  to  prove  two  triangles  congruent : 

1.  Show  that  they  have  the  relation  a.  s.  a. 

2.  Show  that  they  have  the  relation  s.  a.  s. 

NOTE.  It  will  also  be  helpful  to  recall,  at  this  point,  the  cases  of  equal 
angles  treated  previously  (e.g.  right  angles,  vertical  angles,  etc.),  as  well 
as  the  equalities  resulting  from  a  use  of  the  axioms  of  equality  (§  132). 

EXERCISES 

1.  Given,  in  the  annexed  figure,  that  AD  and  CD  bisect  each 
other  at  0.  Q 

To  prove  that  A  C  =  ED. 

Proof.    1.    &AOC  =  &BOD.  s.a.s. 
For  OA  =  OB,          Hyp. 

OC  =  OD,          Hyp. 
and  ZAOC  =  ZBOD  (being  vertical  angles).  & 

2.  .'.  AC  —  BD  (being  homologous  sides  of  congruent  A). 
State  this  exercise  in  the  form  of 

a  proposition. 

2.  Given,  in  the  figure  for  the  pre- 
ceding exercise,  that  0  is  the  mid-point 
of  CD,  and  that  Z  C  =  Z  D.   Prove  that 
AB,  cutting  CD  in  0,  is  bisected  at  0. 

3.  Given,  in  the  annexed  figure,  that 
the  line  AB  bisects  the  angles  CAD 
and  CBD.    Prove  that  Z  C  =  Z  D. 

4.  Given,  in  the  figure  for  the  preceding  exercise,  that  ZCAD  is 
bisected  by  AB,  and  that  AC  =  AD.   Prove  that  BC  =  BD. 


CONGRUENCE 


77 


5.  Given,    in    the    annexed    figure,    that 
Z  in  =  Z  n,  AC  =  AD,  and  that  the  line  CD 
intersects  AB  at  B.    Prove  that  BC  =  BD. 
State  as  a  general  theorem. 

What  angles  also  are  proved  equal  ?  What 
kind  of  angle,  therefore,  is  /.ABC1  What  is 
the  position  of  AB  relative  to  CZ>? 

6.  Given,  in  the  annexed  figure,  that 
A C  =  BD,  and  Z CAB  =  /.DBA.    Prove 
that  BC  =  AD.    What  other  parts   are 
equal  ? 

7.  Given,  in  the  figure  for  Ex.  6,  that 
ZCAB  =  ZDBA,  and  Z  CAD  =  ZDBC. 
Prove  that  BC  =  AD. 

8.  Given,  in  the  annexed  figure,  that 
AB  =  AC,  and  AD  =  AE.     Prove  that 
BE  =  CD.   What  other  parts  are  equal  ? 

9.  Given,  in  the  figure  for  Ex.  8,  that 
AB  =  AC,   and   ZB  =  ZC.    Prove  that 
ZBDC  =  Z  CEB.    What  other  parts  are 
equal  ? 

137.  In  order  to  obtain  a  sufficient  number  of  equal  parts 
of  two  triangles  to  establish  either  of  the  relations  a.  s.  a. 
or  s.  a.  s.,  it  is  sometimes  necessary  first  to  prove  two  other 
triangles  congruent. 

EXAMPLE 

Given,  in  the  annexed  figure,  that 
AC  =  BD,  and  /.CAB  =  Z^DBA. 

To  prove  that    CO  =  DO. 

Analysis.  CO  and  OD  are  parts  of 
A  CO  A  and  DOB  respectively.  In 
these  triangles  AC  —  BD  (hyp.),  and 
/-x  =  /.y  (vert.  A),  but  these  equali- 
ties are  insufficient  to  establish  the  congruence  of  the  triangles 
by  either  of  the  methods  given.  It  is  therefore  necessary  to 


78 


PLANE  GEOMETRY— BOOK  I 


prove  two  other  triangles  congruent  in  order  to  obtain  a  larger 
number  of  equal  parts.  Triangles  ABC  and  ABD  are  chosen, 
since  they  are  readily  proved  to  be  G 
congruent.  Hence  the  following  : 

Proof.  1.    AABC  =  AABD:   s.a.  s. 
For  AC  =  BD,  Hyp. 

AB  is  common, 
and  ^CAB  =  Z.DBA.     Hyp.  ^  * 

2.  .*.  ZC  =  Z/),  and  Z.p  —  Z.q.  Why? 

3.  Then  /.CAB  —  /.p  =  /.DBA  —  Z.q.  Ax.  3 
That  is,                             /-in  =  /.n. 

4.  .'.  ACOA  =  ADOB.  a.  s.  a. 

5.  .'.  CO  =  DO.  Why? 


EXERCISES 

1.  Given,  in  Fig.  1,  that  Zm  =  Zra,  and  £x  =  /.y.    Prove  that 
CB  =  BD. 

2.  Given,  in  Fig.  2,  that  AB  =  AC,  and  AD  =  AE.    Prove  that 


B      A 


FIG.  1 


FIG.    3 
=  /.ECB,  and  Zx  =  Zy.    Prove 


3.  Given,  in  Fig.  2,  that 
that  ^£  =  ^4(7. 

4.  Given,  in  Fig.  3,  that  AB  =  AC,  and  AD  =  AE.   Prove  that 
BO  =  CO. 


CONGRUENCE  79 

PROPOSITION  III.    THEOREM 

138.  If  two  sides  of  a  triangle  are  equal,  the  angles  opposite 
these  sides  are  equal. 


4  A' 

A 


\ 
\ 
\ 


Given  the  triangle  ABC,  in  which  AB  equals  AC. 
To  prove  that  Z  B  =  Z  C. 

Proof.  1.  Construct  the  AA'B'C'  congruent  to  A  ABC,  by 
making  A'B'  =  AB,  A'C'  =  AC,  and  Z.A'  =  Z.A.  s.  a.  s. 

2.  Then     Z.B1  may  be  made  to  coincide  with  /.B. 

(Homologous  angles  of  congruent  A.) 

3.  But  AA'B'C'  may  be  made  to  coincide  with  A  ABC,  even 
when  turned  over  and  placed  so  that  B'  falls  on  C  and  C'  falls 
on  B.  s.  a.  s. 

4.  Then  Z.B'  will  coincide  with  Z  C.  Why  ? 

5.  /.Z£  =  ZC.  Ax.  1 

(Each  being  equal  to  Z.B'.) 

REMARK.  Proposition  III  may  also  be  stated  thus :  In  an  isosceles 
triangle  the  angles  opposite  the  equal  sides  are  equal. 

139.  COROLLARY.*  An  equilateral  triangle  is  also  equi- 
angular. 

For  any  side  may  be  regarded  as  the  base  of  an  isosceles  triangle. 

REMARK.  Note  that  Proposition  III  may  be  proved  by  applying  the 
method  of  Ex.  2,  p.  78.  This  is  similar  to  the  method  of  Euclid,  Book  I, 
Proposition  V  (the  pons  asinorum). 

*  A  corollary  is  a  truth  readily  deduced  from  a  statement  immedi- 
ately preceding. 


80  PLANE  GEOMETRY— BOOK  I 

PROPOSITION  IV.    THEOREM 

140.  If  two  angles  of  a  triangle  are  equal,  the  sides  oppo- 
site those  angles  are  equal. 

A  A' 


\ 


A 

/  \ 

/  \ 

/  \ 

B'L \Q, 

Given  the  triangle  ABC,  in  which  angle  B  equals  angle  C. 
To  prove  that  AB  =  AC. 

Proof.    1.  Construct  the  AA'B'C'  congruent  to  A  ABC,  by 
making  B'C'  =  BC,  Z.B'  =  /.B,  and  ZC"  =  ZC.  a.  s.  a. 

2.  Then    A'B'  may  be  made  to  coincide  with  AB. 

(Homologous  sides  of  congruent  &.) 

3.  But  AA'B'C'  may  be  made  to  coincide  with  A  ABC,  even 
when  turned  over  and  placed  so  that  B'  falls  on  C  and  C'  falls 
on  B.  a.  s.  a. 

4.  Then  A'B'  will  coincide  with  AC.  Why  ? 

5.  .'.  AB  =  AC.  Ax.  1 
(Each  being  equal  to  A'B'.) 

141.  COROLLARY.    An   equiangular    triangle    is    also    equi- 
lateral. 

Proposition  IV  is  the  converse  of  Proposition  III. 
The  converse  of  a  theorem  is  another  theorem  which  results  from  in- 
terchanging the  hypothesis  and  the  conclusion  of  the  given  theorem. 
Thus  in  Proposition  III  the 

hypothesis  is,  two  sides  of  a  triangle  are  equal ; 

conclusion  is,  the  angles  opposite  those  sides  are  equal. 
While  in  Proposition  IV  the 

hypothesis  is,  two  angles  of  a  triangle  are  equal ; 

conclusion  is,  the  sides  opposite  those  angles  are  equal. 
The  converse  of  a  proposition  is  not  necessarily  true,  but  must  be  proved. 


CONGRUENCE 


81 


PROPOSITION  V.    THEOREM 

THIRD    TRIANGLE    CONGRUENCE 


142.  If  two  triangles  have  the  three  sides  of  one  equal 
respectively  to  the  three  sides  of  the  other,  the  triangles  are 
congruent,  (s.  s.  s.) 


Given  the  triangles  ABC  and  A'B'C',  in  which  AB  equals  A'B', 
BC  equals  B'C1,  and  AC  equals  A'C'. 

To  prove  that  A  ABC  =  AA'B'C'. 

Proof.    1.  On  that  side  of  AC  which  is  remote  from  B  con- 
struct A  ADC  congruent  to  AA'B'C'.  a.  s.  a. 

(AC  =  A'C',  Z.CAD  -  Z.A',  and  Z.ACD  -  ZC".) 
Draw  BD. 

2.  Then  Z  x  =  Z  x1,  and  Zy  =  Z.y'.  §  138 


(AB  =  AD,  and  CB  -  CD,  by  hyp.  and  cons.) 

3.  .'./.ABC  =  Z.ADC. 

4.  .'.AABC   =  AADC. 

/.  A.47JC  =  AA'B'C1. 


Ax.  2 

s.  a.  s. 
§  113 


Discussion.  Why  cannot  this  proposition  be  proved  directly  by  super- 
position ?  Note  that  in  the  figure  it  is  tacitly  assumed  that  ZBAC  and 
Z.BCA  are  both  acute.  What  change  would  be  necessary  in  the  figure 
and  proof  if  either  of  these  angles  were  obtuse  ?  Would  a  proof  be 
necessary  if  either  of  these  angles  were  a  right  angle  ? 


82  PLANE  GEOMETRY—  BOOK  I 

143.  Propositions  III,  IV,  and  V  enable  us  to  enlarge  the 
scheme  of  §  136  as  follows  : 


In  order  to  prove  two     ajgs    equal  : 

I.  Show  that  they  are  homologous  t  S1  ®  S  j  of  congruent  A. 

In  order  to  prove  two  A  congruent  : 

1.  Show  that  they  have  the  relation  a.  s.  a. 

2.  Show  that  they  have  the  relation  s.  a.  s. 

3.  Show  that  they  have  the  relation  s.  s.  s. 

H.  Show  that  in  a  A  they  are 


EXERCISES 

1.  Given,  in  the  quadrilateral  A  BCD, 
AD  =  BC,   AB  =  CD.    Prove  Z.I  =  ZC. 
What    construction    line    must    first    be 
drawn  ? 

What  other  angles  may  also  be  proved 
equal  ?   State  as  a  general  theorem.  FIG.  1 

2.  Given,   in  the  isosceles  triangle  ABC  (Fig.  2  below),   that 
BD  =  CE.    Prove  that  AD  —  AE.    State  as  a  general  theorem. 

3.  Given,  in  Fig.  3  below,  that  AB  =  CD,  AD  =  BC,  BO  =  OD. 
Prove  that  EO  =  OF.    State  as  a  general  theorem. 


D        E 
FIG.  2 


4.  Prove  that  if  the  diagonals  of  a  quadrilateral  bisect  each  other, 
the  opposite  sides  of  the  quadrilateral  are  equal. 

5.  Prove  that  if  the  opposite  sides  of  a  quadrilateral  are  equal,  the 
diagonals  bisect  each  other. 


CONGRUENCE 


83 


D. 


6.  Given,  in  Fig.  1,  that  A  ABC  is  equilateral,  and  that  AD  =  BE 
=  CF.   Prove  that  A  DBF  is  also  equilateral. 

A. 

7.  Given,  in  Fig.  2  below,  that  A  ABC  is  equi- 
lateral, with  its  sides  produced  in  succession  so 
that  AD  =  BE  =  CF.    Prove  that  &DEF  is  also 
equilateral. 

8.  Given,  in  Fig.  3  below,  that  A  ABC  is  equi- 
lateral,   and    that   Zm  =  Zn  =  Zo.    Prove   that 
A  DEF  is  also  equilateral. 

9.  Given,  in  Fig.  4  below,  that  A  ABC  is  equilateral,  and  that 
AF  =  BD  =  CE.   Prove  that  &HKL  is  also  equilateral. 


B 


E 
FIG.  1 


C 


D 


0 


FIG.  3 


FIG.  4 


144.  Congruence  of  Polygons.  The  congruence  of  two  poly- 
gons can  be  proved  by  methods  similar  to  those  of  Proposi- 
tions I  and  II.  A  polygon  of  four  sides  has  eight  parts,  one 
of  five  sides  ten  parts,  and,  in  general,  one  of  n  sides  has  2  n 
parts,  n  sides  and  n  angles.  It  can  be  proved  by  superposition 
that  if  two  polygons  of  n  sides  have  2  n  —  3  consecutive  parts 
of  one  respectively  equal  to  2  n  —  3  consecutive  parts  of  the 
other,  the  polygons  are  congruent.  The  abbreviations  of  Prop- 
ositions I  and  II  may  be  extended  to  indicate  the  congruence 
-of  two  polygons,  thus  : 

Two  quadrilaterals  are  congruent  if  they  have  the  relation 
a.  s.  a.  s.  a.,  or  the  relation  s.  a.  s.  a.  s. 

EXERCISES 

1.  In  Ex.  3,  p.  82,  prove    that   the  quadrilaterals  ABOE  and 
CDOF  are  congruent. 

2.  In  Fig.  4,  above,  name  all  congruent  quadrilaterals. 


84  PLANE  GEOMETRY— BOOK  I 

CONSTRUCTIONS 

145.  In  order  to  show  the  equality  of  two  lines  or  two  angles 
in  a  figure,  it  is  often  necessary  to  introduce  additional  elements 
(points,  lines,  and  circles)  in  such  a  way  as  to  form  two  tri- 
angles which  may  be  proved  congruent. 

The  process  by  which  these  elements  are  introduced  into  a 
figure  is  called  construction. 

146.  There  are  in  plane  geometry  three*  fundamental  con- 
structions upon  which  all  other  processes  of  construction  are 
based.    These  have  already  been  used  repeatedly.    They  may 
be  stated  as  follows  : 

1.  Joining  any  two  points  by  a  straight  line. 

2.  Producing  any  straight  line  (segment)  indefinitely. 

3.  Describing  a  circle  about  any  point  as  a  center,  and  with 
a  radius  equal  to  a  given  line  (segment). 

147.  These  constructions  often  make  possible  the  determi- 
nation of  points  in  a  figure 

1.  As  the  intersection  of  two  lines  (one  point). 

2.  As  the  intersection  of  a  line  and  a  circle  (two  points). 

3.  As  the  intersection  of  two  circles  (two  points). 

148.  Several  constructions  have  been  shown  to  result  directly: 

1.  Laying  off  on  a  straight  line  a  length  equal  to  a  given 
line  (segment). 

2.  Drawing  a  line  so  that  it  will  form  with  a  given  line,  at 
a  given  point,  an  angle  equal  to  a  given  angle  (§  89). 

3.  The  three  fundamental  triangle  constructions  (§§  90,91,74). 

149.  In  establishing  a  construction  it  is  necessary  not  only 
to  describe  the  process  in  detail,  but  also  to  justify  it ;  that  is,  to 
demonstrate  that  the  steps  in  the  process  lead  to  the  desired  result. 

The  next  theorem  is  of  value  in  proving  some  of  the  most 
important  constructions. 

*  Since  the  time  of  Plato  (about  390  B.C.)  it  has  been  customary  in  ele- 
mentary geometry  not  to  allow  any  other  instruments  than  the  straight- 
edge and  the  compasses. 


THE  KITE 
PROPOSITION  VI.    THEOREM 


85 


150.  If  two  isosceles  triangles  are  constructed  on  the  same 
base,  the  line  that  joins  their  vertices  bisects  the  common  base 
at  right  angles. 

C 


Given  two  isosceles  triangles  ABC  and  ABD,  on  the  common 
base  AB,  and  the  line  CD  cutting  AB  at  E. 
To  prove  that     CE  bisects  AB  at  right  angles. 
Proof.    1.  AACD  =  ABCD. 

AC  =  EC,  and  AD  =  ED, 
CD  is  common. 

.-.Zm  =  Z?i.  Why? 

AACE  =  ABCE.  Why? 

.'.  AE  =  BE, 


For 
and 

2. 

3.  Also 

4. 
and 

5. 


s.  s.  s. 
Hyp. 


Why? 


.'.  CE  bisects  AB  at  right  angles. 
(Each  angle  at  E  being  one  half  of  a  st.  Z.) 

151.  COROLLARY  1.   Two  points  each  equidistant  from  the  ex- 
tremities of  a  line  determine  the  perpendicular  bisector  of  that  line. 

A  quadrilateral  (such  as  A  CBD  in  the  first  of  the  above  figures) 
which  has  two  pairs  of  adjoining  sides  equal  is  called  a  kite. 

152.  COROLLARY  2.   The  diagonals  of  a  kite  are  perpendicular 
to  each  other. 

153.  In  order  to  prove  that  two  lines  are  mutually  _L  : 

1.  Show  that  they  form  equal  supplementary-adjacent  angles. 

2.  Show  that  they  are  diagonals  of  a  kite. 


86  PLANE  GEOMETRY— BOOK  I 

PROPOSITION  VII.    PROBLEM 
154.  To  bisect  a  given  straight  line  (line-segment). 


/ 

\ 

E    1                  7? 

\ 

-" 

/ 

Given  the  line  AB. 
Required  to  bisect  AB. 

Construction.  1.  With  A  and  B  as  centers,  and  with  equal 
radii  sufficiently  great,  describe  arcs  intersecting  at  C  and  D. 
2.  Draw  CD,  cutting  AB  at  E.  Then  E  is  the  mid-point  of  AB. 
Proof.    1.  Draw  A  C,  BC,  AD,  BD. 

(To  be  completed,  §  151.) 

155.  The  line  from  any  vertex  of  a  triangle  to  the  mid-point 
of  the  opposite  side  is  called  the  median  to  that  side. 

EXERCISES 

1.  Explain  how  to  divide  a  given  line-segment  into  4  (8,  16,  2W) 
equal  parts. 

2.  How  many  medians  has  a  triangle?  Draw  a  scalene  triangle 
(§  75)  and  construct  its  medians. 

3.  Draw    a    scalene    triangle    and    construct    the    perpendicular 
bisectors  of  its  sides. 

4.  Prove  that  the  median  to  the  base  of  an  isosceles  triangle  is 
the  perpendicular  bisector  of  the  base. 

5.  Prove  that  if  the  median  of  a  triangle  is  perpendicular  to  the 
base,  the  triangle  is  isosceles. 

6.  Prove  that  the  medians  to  the  legs  of  an  isosceles  triangle 
are  equal. 

7.  Prove  that  homologous  medians  of  congruent  triangles  are  equal. 


CONSTRUCTIONS  87 

PROPOSITION  VIII.    PROBLEM 
156.   To  Used  a  given  angle. 


Given  the  angle  CAB. 
Required  to  biseot  /-CAB. 

Construction.  1.  With  A  as  a  center,  and  with  any  convenient 
radius,  as  AD,  describe  an  arc  cutting  AC  in  D,  and  AB  in  E. 

2.  With  D  and  E  as  centers,  and  with  equal  radii  sufficiently 
great,  describe  arcs  intersecting  in  F.  Draw  AF.  Then  AF 
bisects  Z.A. 

Proof.    1.  Draw  DF  and  EF. 

(To  be  completed,  §  142.) 

157.  COROLLARY.  To  bisect  a  given  are  of  a  given  circle, 
bisect  the  central  angle  intercepted  by  the  arc.  The  line  which 
bisects  this  angle  will,  if  produced,  bisect  the  given  arc  also. 

Why? 

EXERCISES 

1.  Construct  a  scalene  triangle  and  draw  the  bisectors  of  the  inte- 
rior angles,  producing  each  to  meet  the  opposite  side. 

2.  Prove  that  the  bisector  of  the  vertex  angle  of  an  isosceles 
triangle  bisects  the  base  and  is  perpendicular  to  it. 

3.  Prove  that  the  bisectors  of  the  base  angles  of  an  isosceles 
triangle,  terminating  in  the  opposite  sides,  are  equal. 

4.  How  can  an  angle  be  bisected  by  the  use  of  a  carpenter's  square  ? 

5.  Prove  that  two  triangles  are  congruent  if  they  have  a  side, 
an  adjoining  angle,  and  the  bisector  of  that  angle  in  one  equal 
respectively  to  the  corresponding  parts  in  the  other. 


88 


PLANE  GEOMETRY—  BOOK  I 
PROPOSITION  IX.    PROBLEM 


158.  At  a  given  point  in  a  given  line,  to  erect  a  perpen 
dicular  to  that  line. 


\ 


D          C          E 

Given  the  point  C  in  the  line  AB. 
Required  to  erect  a  perpendicular  to  AB  at  C. 
Construction.    1.  Lay  off  CD  =  CE. 

2.  With  D  and  E  as  centers,  and  with  equal  radii  sufficiently 
great,  describe  arcs  intersecting  at  F.    Draw  FC.    Then  FC 
is  the  perpendicular  required. 
Proof.    Draw  FD  and  FE. 

(To  be  completed,  §  142.) 

EXERCISES 

1.  Draw  two  mutually  perpendicular  lines.  On  the  four  rays  lay 
off  equal  segments  from  the  point  of  intersection,  and  connect  their 
extremities.  Prove  that  the  sides  and  the  angles  of  the  resulting 
quadrilateral  are  equal. 


2.  Bisect  a  straight  angle  and  bisect  both  halves,  producing  each 
bisector  in  turn  through  the  vertex.  Then  proceed  as  in  Ex.  1. 
What  kind  of  polygon  is  formed  ?  (See  §  57.) 


CONSTRUCTIONS  89 

PROPOSITION  X.    PROBLEM 

159.  From  a  given  point  without  a  given  line,  to  let  fall  a 
perpendicular  upon  that  line. 

\0 


Given  the  line  AB  and  the  point  C  without  it. 

Required  to  let  fall  a  perpendicular  from  C  to  AB. 

Construction.  1.  With  C  as  a  center,  and  with  a  radius 
sufficiently  great,  describe  an  arc  cutting  AB  in  D  and  E. 

2.  With  D  and  E  as  centers,  and  with  equal  radii  suffi- 
ciently great,  describe  two  arcs  intersecting  at  F.  Draw  CF 
(producing  it  if  necessary)  cutting  AB  in  G.  Then  CG  is  the 
perpendicular  required. 

Proof.    Draw  CD  and  CE,  also  FD  and  FE. 
(To  be  completed,  §  150.) 

160.  An  altitude  of  a  triangle  is  a  perpendicular  let  fall  from 
any  vertex  to  the  side  opposite,  produced  if  necessary.  The 
opposite  side  is  called  the  corresponding  base. 

EXERCISES 

1.  How  many  altitudes  has  a  triangle? 

2.  Construct  the  altitudes  of  a  scalene  acute  triangle ;  of  an  equi- 
lateral triangle ;  of  an  isosceles  acute  triangle. 

3.  Construct  the  altitudes  of  an  obtuse  triangle.    What  additional 
construction  lines  are  necessary  ? 

NOTE.  The  construction  given  above  is  strictly  geometric  (see  footnote, 
p.  84).  In  practice  a  perpendicular  is  drawn  by  the  use  of  a  model  of 
a  right  angle,  such  as  a  carpenter's  square  or  a  draftsman's  triangle 
(see  p.  103). 


90  PLANE  GEOMETRY— BOOK  I 

PROPOSITION  XL    THEOREM 

161.   Of  all  lines  that  can  be  drawn  from  a  given  point  to  a 
given  line,  only  one  is  perpendicular  to  the  given  line. 


Given  the  line  AB,  the  point  C  without  it,  and  CD  perpen- 
dicular to  AB. 

To  prove  that  any  other  line,  as  CF,  drawn  from  C  to  AB  is 
not  perpendicular  to  AB. 

Proof.  1.  Produce  CD  to  E,  making  DE  =  CD,  and  draw  EF. 

2.  Then  ACDF=AEDF.  s.  a.  s. 
For                                 DF  is  common, 

CD  =  DE,  Cons, 

and  Z  m  =  Z  7i.  Why  ? 

3.  .'.Z.p  =  Z.q. 

(Homologous  A  of  congruent  A.) 

4.  Now  CDE  is  a  straight  line.  Cons. 

.'.  CFE  is  not  a  straight  line.  §11 

.'.  Z.CFE  is  not  a  straight  angle. 

5.  .'.Z.p  (one  half  of  Z  CFE)  is  not  a  rt.  Z. 

6.  .'.  CF  is  not  perpendicular  to  AB,  nor  is  any  other  line 
from  C  perpendicular  to  AB  except  CD 

162.  COROLLARY.    A  triangle  can  have  but  one  right  angle. 


BOOK  I  91 

EIGHT  TRIANGLES 
CONSTRUCTION  OF  RIGHT  TRIANGLES 

163.  By  applying  the  methods  of  §§  89,  158,  159  it  is  pos- 
sible to  construct  a  right  triangle,  having  given 

1.  A  leg  and  the  adjoining  oblique  angle. 

2.  The  two  legs. 

3.  The  hypotenuse  and  a  leg. 

4.  The  hypotenuse  and  an  adjoining  angle. 

These  constructions  correspond  to  as  many  special  laws  of 

CONGRUENCE  OF  RIGHT  TRIANGLES 

164.  From  the  First  and  Second  Triangle  Congruences  it 
follows  that 

1.  If  two  right  triangles  have  a  leg  and  the  adjoining  ob- 
lique angle  of  one  equal  respectively  to  a  leg  and  the  adjoining 
oblique  angle  of  the  other,  the  triangles  are  congruent,    (a.  s.  a.) 

2.  If  two  right  triangles  have  two  legs  of  one  equal  respec- 
tively to  two  legs  of  the  other,  they  are  congruent.  (s.  a.  s.) 

EXERCISES 

1.  Prove  that  if  the  altitude  of  a  triangle  bisects  the  base,  the 
triangle  is  isosceles. 

2.  Prove  that  if  the  altitude  of  a  triangle  bisects  the  vertex  angle, 
the  triangle  is  isosceles. 

3.  Roman  surveyors,  called  agrimensores,  are  said  to  have  used 
the  following  method  of  measuring  the  width  of  a  stream :  A  and 
B  were  points  on  opposite  sides  of  the  stream,  in  plain  view  from 
each  other.    The  distance  AD  was  then  taken  at  right  angles  to  AB, 
and  bisected  at  E.  Then  the  distance  DF  was  taken  at  right  angles 
to  AD,  such  that  the  points  B,  E,  and  F  were  in  a  straight  line. 
Make  a  drawing  illustrating  the  above  method,  show  what  meas- 
urement affords  a  solution,  and  prove  that  this  is  so. 


92 


PLANE  GEOMETBY— BOOK  I 


PROPOSITION  XII.    THEOREM 

165.  If  two  right  triangles  have  the  hypotenuse  and  a  leg 
of  one  equal  respectively  to  the  hypotenuse  and  a  leg  of  the 
other,  the  triangles  are  congruent,  (rt.  A  h.  1.) 


Given  the  right  triangles  ABC  and  A'B'C',  with  the  hypotenuse 
AB  equal  to  the  hypotenuse  A'B',  and  with  AC  equal  to  A'C*. 

To  prove  that  rt.  A  ABC  =  rt.  A  A'B'C1. 

Proof.   1.  On  AB  (  =  A'B')  construct  rt.  AADB  congruent  to 
rt.  A  A'B'C',  so  that  AD  =  A'C1,  and  BD  =  B'C'.    Draw  CD. 


2.  Then 

3.  Now 

and  hence 

4. 
that  is, 


(Since  AC  =  AD,  by  hyp.  and  cons.) 

/LACB=^ADB. 


EC  =  BD. 
.'.  AACB  =  AADB, 
A  ABC  =  A  A'B'C'. 


§138 

Why? 

Ax.  3 

Why  ? 

s.  s.  s. 

Why  ? 


EXERCISES 

1.  Prove  that  a  ladder  of  known  length,  the  foot  of  which  is  at 
a  known  horizontal  distance  from  the  wall  of  a  house,  always  reaches 
the  same  distance  up  the  side  of  the  house. 

2.  Prove  that  the  altitude  to  the  base  of  an  isosceles  triangle 
bisects  the  base  and  the  vertex  angle. 


EIGHT  TKIAKGLES  93 

PROPOSITION  XIII.    THEOREM 

166.  If  two  right  triangles  have  the  hypotenuse  and  an  adjoin- 
ing angle  of  one  equal  respectively  to  the  hypotenuse  and  an  ad- 
joining angle  of  the  other,  the  triangles  are  congruent,  (rt.  A  h.  a.) 
B  B' 


D  AC'  A 

Given  the  right  triangles  ABC  and  A'B'C',  with  the  hypotenuse 

AB  equal  to  the  hypotenuse  A  'B',  and  with  the  angle  A  equal  to 

the  angle  A'. 

To  prove  that       rt.  A  ABC  =  rt.  AA'B'C'. 

Proof.    1.  On  AC  take  AT)  =  A'C',  and  draw  BD. 

2.  Then  AABD  =  AA'B'C1.  s.  a.  s. 

3.  Hence  /.ADE  (=  Z.A'C'B')  is  a  rt.  Z.         Why? 

4.  .\BD  coincides  with  BC.  §  161 

5.  .:AABD  =  AABC;  Def. 
that  is,                            AABC  =  AA'B'C'.                            Why? 

167.  A  further  enlargement  of  the  scheme  of  §  143  is  now 
possible. 

In  order  to  prove  two  L  equal  : 


I.  Show  that  they  are  homologous  (       .     J  of  congruent  A. 

In  order  to  prove  two  triangles  congruent  : 

1.  Show  that  they  have  the  relation  a.  s.  a. 

2.  Show  that  they  have  the  relation  s.  a.  s. 

3.  Show  that  they  have  the  relation  s.  s.  s. 

4.  Show  that  they  have  the  relation  rt.  A  h.  1. 

5.  Show  that  they  have  the  relation  rt.  A  h.  a. 

__    _     •  .   .  /sides  opposite  equal  anglesX 

II.  Show  that  in  a  A  they  are  (angles  ^posite  4equal 


94  PLANE  GEOMETRY— BOOK  I 

EXERCISES 

1.  Prove  that  the  perpendiculars  let  fall  from  the  mid-points  of 
the  legs  of  an  isosceles  triangle  upon  the  base  are  equal. 

2.  Prove  that  the  altitudes  on  the  legs  of  an  isosceles  triangle 
are  equal. 

3.  Prove  that  homologous  altitudes  of   congruent  triangles   are 
equal. 

4.  Prove  that  the  bisectors  of  homologous  angles  of  congruent 
triangles  are  equal. 

5.  In  transferring  a  line-segment  by  means  of  the  dividers,  which 
law  of  congruence  is  applied?    Is  it  necessary  that  the  legs  of  the 
dividers  should  be  of  equal  length  ? 

6.  By  what  measurements  can  a  carpenter  ascertain  that  the  two 
gable  triangles  of  a  roof  are  exactly  alike  ? 

7.  Show  how,  by  measuring  sides  and  diagonals,  a  surveyor  may 
determine  completely  the  size  and  shape  of  an  irregular  field,  with- 
out measuring  any  angles.    (Assume  that  the  boundaries  of  the 
field  are  straight  lines.) 

8.  State  the  above  problem  in  the  form  of  a  geometric  theorem 
in  the  comparison  of  polygons. 

PARALLEL  LINES 

168.  A  line  cutting  two  or  more  lines  is  a  transversal  of 
these  lines.    The  transversal  t 
in  the  figure   forms  with  the 
lines  m  and  n  angles  that  are 
classified  as  follows : 

a,b,c',d'  are  interior  angles ; 

a',b',c,d  are  exterior  angles  ; 

a  and  c'  are  alternate-interior 
angles,  also  b  and  d';  c/d 

a'  and  c  are  alternate-exterior 
angles,  also  I'  and  d ; 

a'  and  a  are  exterior-interior  or  corresponding  angles,  also  b' 
and  b,  c'  and  c,  d '  and  d. 


PAEALLEL  LINES  95 

PROPOSITION  XIV.    THEOREM 

169.  When  two  straight  lines  are  cut  by  a  transversal,  if 
the  alternate-interior  angles  are  equal,  the  two  straight  lines 
do  not  intersect. 


Given  two  straight  lines  AB  and  CD  cut  by  the  transversal  RS, 
with  the  angle  AEF  equal  to  the  angle  EFD. 

To  prove  that  AB  cannot  intersect  CD. 

Proof.  1.  If  possible,  let  AB  intersect  CD  at  L.  Take  FM  = 
EL,  and  draw  EM. 

2.  Then                       AEFL  =  AEFM.  s.  a.  s. 
For                                     EL  =  FM,  Cons. 

EF  is  common, 
and  Z  EFM  =  Z  PEL.  Hyp. 

3.  /.  Z EFL  =  Z  FEM.  Why  ? 

4.  But  Z.EFL  +  Z.EFM  =  a  st.  Z.  Why  ? 

5.  Hence  Z.FEM  +  Z  F£L  =  a  st.  Z.  Ax.  1 

6.  Hence  LEM  would  be  a  straight  line  cutting  CD  in  two 
points  Z  and  M. 

7.  Since  that  is  impossible  (why  ?), 

AB  cannot  intersect  CD. 
(The  method  of  proof  used  here  is  called  the  Indirect  Method.) 

170.  Parallel  lines  are  lines  which  lie  in  the  same  plane  and 
do  not  meet,  however  far  they  are  produced. 

Thus  in  the  above  theorem  AB  is  parallel  to  CD.  This  is 
often  written  AB  \\  CD. 


96  PLANE  GEOMETRY— BOOK  I 

Proposition  XIV  may  now  be  stated :  When  two  straight 
lines  are  cut  by  a  transversal,  if  the  alternate-interior  angles  are 
equal ,  the  straight  lines  are  parallel. 

The  figures  below  show  that  if  the  alternate-interior  angles 
are  equal,  then  the  alternate-exterior  angles  are  also  equal ;  and 
if  the  corresponding  angles  are  equal,  the  two  sets  of  alternate 
angles  are  equal,  etc.  In  fact,  if  any  one  of  the  following 
twelve  equations  is  true,  they  are  all  true. 

i50°/30°  b'/af 


30150°  C/d 


C/ 


Z  b'  =  Z  d    alternate  Z  I  =  Z  V    corresponding 

Z  a  =  Z  c1  I    angles.  Z  c  =  Z  c'  f       angles. 

\ 


Z  a'  +  Z  d  =  a  st.  Z.  Z  a  +  Z  d'  =  a  st.  Z. 

Z  V  +  Z  c  =  a  st.  Z.  Z  6  +  Z  c'  =  a  st.  Z. 

PROPOSITION  XV.    THEOREM 

171.  When  two  straight  lines  are  cut  by  a  transversal,  if 

(a)  the  alternate-exterior  angles  are  equal  ;  or  if 

(b)  the  exterior-interior  angles  are  equal  ;  or  if 

(c)  an'y  two  interior  or  any  two  exterior  angles  on  the 

same  side  of  the  transversal  are  supplementary, 
the  two  straight  lines  are  parallel. 

For  any  one  of  these  relations  may  be  reduced  to  the  equality 
of  alternate-interior  angles. 

172.  COROLLARY.    Two  straight  lines  in  the  same  plane  per- 
pendicular to  the  same  straight  line  are  parallel. 


PAEALLEL  LINES  97 

PROPOSITION  XVI.    PROBLEM 

173.   To  construct  a  line  parallel  to  a  given  line,  through  a 
given  external  point. 


E 


Given  a  straight  line  AB  and  the  external  point  P. 
Required  to  construct  through  P  a  line  parallel  to  AB. 

Construction.   1.  Through  P  draw  a  transversal  cutting  A  B  at  D. 
2.  Construct  Z  cW  =  Z  CDB.    Then  the  line  EPF  is  II  to 
AB.  \  §  171 

(What  other  angle-pairs  might  have  been  used  for  this  construction  ?) 


174.  We  now  assulne  .the  following 

Parallel  Axiom.    Thrbufrh  a  given  point  but  one  straight  line 
can  be  drawn  parallel  t\  a  given  'straight  line. 

This  assumption  may  blso  be  stated  as  follows: 
Two  intersecting  straight  lines  cannot  both  be  parallel  to  the 
same  straight  line. 


175.  COROLLARY.  :T4w?  straight  lines  in  the  same  plane  par- 
allel to  the  same  straight  line  are  parallel  to  each  other. 

EXERCISE 

1.  Given  in  the  annexed  figure  that  Z  c  —  Z  a  +  Z  b.    Prove  that 
l,\\ly  I, 

Suggestion.  Through  the  vertex  of  Z  c  draw  a  \# 
line  /3  dividing  Z  c  into  Z  x  and  Z  y  and  mak-  c/> 
ing  Z  x  -  Z  a.  Then  ^  II  /3.  (Why  ?)  Also  /2  II  /3.  /}> 
(Why?)  ,.1,111,.  (Why?)  IT" 


98  PLANE  GEOMETRY— BOOK  I 

PROPOSITION  XVII.    THEOREM 

176.  If  two  parallel  lines  are  cut  by  a  transversal : 

(a)  the  alternate-interior  angles  are  equal; 

(b)  the  exterior-interior  angles  are  equal; 

(c)  the  alternate-exterior  angles  are  equal; 

(d)  any  two  interior  or  any  two  exterior  angles  on  the  same 
side  of  the  transversal  are  supplementary. 


M rr 

fl. 


N 


F 

Given  the  parallel  lines  AB  and  CD  cut  by  the  transversal  EF 
at  H  and  K. 

To  prove  (a)  that       Z  A  HK  =  Z  HKD. 

Proof.    1.  If  ^LAHK is  not  equal  to  Z.HKD,  draw  MN through 
H,  making  /.MHK  =  ^HKD. 

2.  Then  MN  II  CD.  §  170 

(Since  the  alt. -int.  A  are  equal.) 

3.  But  AB  II  CD.  Hyp. 
Then  MN  and  AB  are  two  lines  drawn  through  the  same 

point  H,  and  parallel  to  CD. 

4.  But  this  is  impossible.  §  174 

5.  Hence  MN  must  coincide  with  A  B. 

.'.  ^MHK  =  Z.AHK  =  Z.HKD.  Why  ? 

Conclusions  (b),  (c),  (d)  follow  directly  from  this. 

(To  be  completed.) 
(This  theorem  is  the  converse  of  Proposition  XV.) 


PARALLEL  LINES 


99 


177.  COROLLARY  1.   A  straight  line  perpendicular  to  one  of 
two  parallel  lines  is  perpendicular  to  the  other  also. 

178.  COROLLARY  2.    If  two 
lines  are  perpendicular  respec- 
tively to  two  intersecting  lines, 
they  cannot  be  parallel. 

If  Pj-Lm,  and_p2_Ln,  then  Za<rt.Z, 
and  Z  6<rt.  Z.  .-.  pl  is  not  parallel 
to  p2,  since  Z  a  +  Z  b  <  a  st.  Z. 

179.  COROLLARY  3.    If  two  angles  have  their  sides  respec- 
tively parallel,  they  are  equal  or  supplementary. 

The  following  diagrams  illustrate  various  relative  positions  of  such 
angles. 


A 


V 


EXERCISES 

1.  Give  ten  concrete  illustrations  of  parallels. 

2.  In  the  figure,  m\\n,  and  one  of  the       


exterior  angles  is  30°.    Find  the  values 
of  the  other  angles. 

3.  Howmany  sets  of  numerically  differ-      _J1 
ent  angles  does  the  figure  (Ex.2)  contain ?    /^ 

4.  If  the  exterior  angles  (Ex.  2)  are  in  the  ratio  1 :  2  (2  :  3  ;  4:5), 
determine  their  values. 


100 


PLANE  GEOMETKY— BOOK  I 


5.  If  the  ratio  of  the  angles  in  Ex.  4  is  1:1,  what  is  the  position 
of  the  transversal  with  reference  to  the  parallels  ? 

6.  Let  the  figure  represent  two  inter- 
secting streets.     How    many   numerically 
different  angles  are  there  ? 

7.  If  one  of  the  streets  extends  in  an 
east-and-west  direction,   and  the  ratio  of 
the  angles  is  1 : 3,  what  is  the  direction  of 
the  other  street? 

8.  The  T-square  is  an  instrument  constantly  used  in  mechanical 
drawing  (see  illustration).   By  its 

means  a  series  of  parallel  lines 
may  be  drawn  at  any  desired 
intervals.  Upon  what  theorem  of 
parallels  does  its  use  depend? 

9.  Copy  the  following  Greek  designs,  called  meanders.    What 
method  of  constructing  parallels  is  used  ? 


PARALLEL  LINES 


FIG.  1 


FIG.  2 


101 


10.  Which    laws    of    parallels    are    illustrated "  by Hh^   follcwirg? 
capital  letters:  Z,  N,  H,  E,  F,  W  ? 

11.  In  Fig.  1,  ^  II 12  ;  prove  that  Zc  =  Za  +  Zb. 


12.  In  Fig.  2,  /x  II  /2  ;  prove  that  Z.  a  +  /_  I  +  Z.  c  =  2  st.  A. 

13.  If  7j,  /2,  etc.  are  different  lines,  construct  the  following  figures  : 


(4) 


Jo  II  L 


/J_/ 


Z2  II  /8, 
Z3±Z4 


In  (1)  what  is  the  position  of  Z2  with  reference  to  /3  ?  in  (2)-(6) 
what  are  the  relative  positions  of  the  first  line  and  the  last  line  ? 

14.  Construct  three  equilateral  triangles  in 
the  positions  shown  in  the  figure.    Prove  that 
I II  a.   What  other  lines  are  parallel  ? 

15.  The  following  is  a  practical  method  of 
drawing  through  a  given  point  a  line  parallel 

to  a  given  line.    Perform  the  construction  as  indicated,  and  prove 
that  it  leads  to  the  desired  result. 

Given  the  line  BC  and  the  point  A.    Required  to  draw  through 
A  a  line  parallel  to  BC. 

With  A  as  a  center  and  a  radius  sufficiently  great  describe  a  circle 
cutting  the  line  BC  in  the  point  D.  Draw  AD.  On  BC  lay  off  DE 
equal  to  AD,  and  in  the  circle  draw  a  chord  DF  equal  to  the  dis- 
tance AE,  so  that  DF&nd  AE  lie  on  the  same  side 
of  BC,  but  on  opposite  sides  of  AD.  Draw  AF  and 
produce  it  if  necessary.  AF  is  parallel  to  BC. 

16.  In  the  accompanying  figure  there  are  four 
lines  which  are  drawn  parallel  to  the  diagonal  of 
the  square.    Do  they  appear  to  be  parallel  to  the 
diagonal  and  to  one  another  ? 


102  PtANE  GEOMETRY— BOOK  I 

ANGLE-SUM 
PROPOSITION  XVIII.    THEOREM 

180.    The  sum  of  the  interior  angles  of  a  triangle  equals  a 
straight  angle. 

E  B 


Given  the  triangle  ABC,  whose  angles  are  m,  n,  and  o. 
To  prove  that          Z  m  -f-  Z  n  +  Z  o  =  a  st.  Z. 
Proof.    1.  Through  B  draw  EF  II  A  C. 

2.  Then  Zw  =  Zj9,  §  176 
and                                            Z  n  =  Z  q.                                Why  ? 

3.  /.  Zra-f-Zo +  Ztt  =  Z^ +  Zo +  Z0;.          Ax.  2 

4.  But          Z_p  +  Z0  +  Z?  =  a  st.  Z.  Why  ? 

. ' .  Z  m  -f-  Z  o  •+•  Z  ft  =  a  st.  Z.  Ax.  1 

181.  COROLLARY  1.    An  exterior  angle  of  a  triangle  is  equal 
to  the  sum  of  the  two  remote  interior  angles,  and  is  therefore 
greater  than  either  of  them. 

182.  COROLLARY  2.    If  two  angles  of  one  triangle  are  equal 
respectively  to  two  angles  of  another  triangle,  the  third  angles 
are  equal. 

183.  COROLLARY  3.    Two  triangles  are  congruent  if  a  side 
and  any  two  angles  of  one  are  equal  respectively  to  a  side  and 
two  angles,  similarly  situated,  of  the  other. 

184.  COROLLARY  4.    The  sum  of  two  angles  of  a  triangle  is 
less  than  a  straight  angle. 

185.  COROLLARY  5.    If  a  triangle  has  one  right  angle  or  one 
obtuse  angle,  the  other  angles  are  acute. 


ANGLE-SUM  103 

186.  COROLLARY  6.    Every  triangle  has  at  least  two  acute 
angles. 

187.  COROLLARY  7.    The  sum  of  the  two  acute  angles  of  a 
right  triangle  equals  a  right  angle. 

188.  COROLLARY  8.    Each  angle  of  an  equilateral  triangle  is 
an  angle  of  60°. 

189.  COROLLARY  9.    If  one  leg  of  a  right  triangle  is  half  the 
hypotenuse,  then  the  angle  opposite  that  leg  is  an  angle  of  30°, 
and  the  other  acute  angle  is  an  angle  of  60°. 

190.  COROLLARY  10.    If  a  right  triangle  contains  acute  angles 
of  80°  and  60°  respectively,  then  the  leg  opposite  the  angle  of  30° 
is  one  half  the  hypotenuse. 

191.  COROLLARY  11.    If  the  legs  of  a  right  triangle  are  equal, 
each  acute  angle  is  an  angle  of  45°. 

192.  COROLLARY  12.    If  one  acute  angle  of  a  right  triangle 
is  an  angle  of  45°,  then  the  other  acute  angle  is  also  an  angle  of 
45°,  and  the  legs  of  the  right  triangle  are  equal. 

193.  COROLLARY  13.    Each  base  angle  of  an  isosceles  triangle 
is  half  the  supplement  of  the  vertex  angle. 


1.  In  mechanical  drawing  constant  use  is  made  of  certain  fixed 
right  triangles  which  are  made  of  wood  or  celluloid.    One  of  these 
has  acute  angles  of  30°  and  60°,  while 

the  other  has  acute  angles  of  45° 
each.  What  angles  can  a  draftsman 
draw  with  the  aid  of  these  two  tri- 
angles, but  without  bisecting  any 
angle  ? 

2.  Show  how  one  of  the  above  tri- 
angles, combined  either  with  the  other  triangle  or  with  the  T-square 
(see  Ex.  8,  p.  100),  may  be  used  to  draw  a  series  of  parallels. 


104  PLANE  GEOMETRY— BOOK  I 

EXERCISES 

1.  Prove  the  proposition  concerning  the  sum  of  the  angles  of  a 
triangle  by  using  each  of  the  following  diagrams : 


2.  Can  a  triangle  be  formed  with  the  following  angles :  (1)  40°, 
50°,  90°?  (2)  30°,  20°,  50°?  (3)  70°,  80°,  100°?  (4)  100°,  200°,  300°? 
(5)  10°,  20°,  30°?  (6)  75 -J°,  4-|0,  100£°?  (7)  (90  -  a)°,  (90  +  a)°,  6°? 
(8)  (90-a?)°,(90  +  y)°,(a:-y)0?  (9)  76°  31' 10",  94°  18'  6",  9°10'45"? 

3.  Find  the  third  angle  c,  if  two  angles  of  a  triangle,  a  and  b, 
have  the  following  values  : 

(1)  (2)  (3)  (4)  (5)  (6) 

«  =  20°  60°  100°  45°  72^°  72° 

&  =  30°  70°  79°  45°  45°  36° 

4.  Determine  the  values  of  the  angles  a,  b,  c  of  a  triangle,  if  (1) 
their  ratio  is  1 :  2  :  3  ;  (2)  a  =  b,  b  =  \  c ;  (3)  a  =  b  +  20°,  b  =  c  +  20°; 
(4)  a  +  b  =  70°,  b  +  c  =  150° ;   (5)  their  ratio  is  1 : 1 :  2. 

5.  Find  the  value  of  each  angle  of  an  isosceles  triangle,  if  the 
vertex  angle  contains  30°;  40°;  50°;  f  rt.Z;  60°;  70°;  £  ft.  Z;  80°; 
100°;  f  rt.  Z;  120°;  130°;  150°;  42°  16';  16°28/52// 

6.  Find  the  value  of  each  angle  of  an  isosceles  triangle,  if  one 
of  the  base  angles  contains  30°;  35°;  40°;  45°;  50°;  f- rt.  Z ;  55°; 
f  st.  Z ;  80° ;  89° ;  57°  43' ;  68°  51'  1". 

7.  One  acute  angle  of  a  right  triangle  is  30° ;  40° ;  45° ;  48-|° ;  60° ; 
62° ;  42  \° ;  22^°.  Find  the  value  of  the  other  acute  angle  and  of  each 
of  the  exterior  angles. 

8.  How  many  degrees  in  each  exterior  angle  of  an  equilateral 
triangle  ? 

9.  In  what  kind  of  triangle  is  one  angle  sufficient  for  the  deter- 
mination of  the  other  two  ? 

10.  If  the  vertex  angle  of  an  isosceles  triangle  contains  120°,  show 
that  the  altitude  divides  the  triangle  into  two  triangles  that  can  be 
placed  so  as  to  form  an  equilateral  triangle. 


ANGLE-SUM 


105 


11.  What  is  the  resulting  figure  in  Ex.  10  if  the  vertex  angle  is  a 
right  angle  ? 

12.  The  base  angles  of  an  isosceles  triangle  are  each  40°.   Find  the 
angle  formed  by  their  bisectors ;  also  the  angle 

formed  by  the  bisectors  of  their  exterior  angles. 

13.  InthefigureZa  +  Z6  =  ast.Z.  What 
kind  of  triangle  is  it  ? 

14.  Each  angle  of  a  triangle  is  the  supple- 
ment of  the  sum  of  the  other  two. 

15.  Two  angles  of  a  triangle  cannot  be  supplementary. 

16.  The  base  angles  of  an  isosceles  triangle  are  acute. 

17.  An  isosceles  triangle  is  obtuse,  right,  or  acute,  according  as  a 
base  angle  is  less  than,  equal  to,  or  greater  than  45° 

18.  If  an  angle  of  an  isosceles  triangle  is  obtuse,  it  is  the  vertex  angle. 

19.  The  sum  of  the  acute  angles  of  an  obtuse  triangle  is  an  acute 
angle. 

20.  The  bisectors  of  the  base  angles  of  an  isosceles  triangle  form 
with  the  base  an  isosceles  triangle  whose  vertex  angle  is  the  supple- 
ment of  each  base  angle  of  the  first  triangle. 

21.  If  the  sides  of  one  angle  are  perpendicular  to  the  sides  of  an- 
other angle,  the  angles  are  equal  or  supplementary. 


22.  If  a  perpendicular  is  dropped  from  the  vertex  of  the  right 
angle  of  a  right  triangle  to  the  hypotenuse,  prove  that  the  two  tri- 
angles thus  formed  and  the  given  triangle  are  mutually  equiangular. 


106 


PLANE  GEOMETRY— BOOK  I 


23.  The  perpendiculars  dropped  from  any  point  in  the  bisector  of 
an  angle  to  its  sides  make  equal  angles  with  the  bisector. 

24.  In  the  first  figure  below  prove  that  ZD  is  greater  than  ZA. 

C 


25.  If  one  of  the  equal  sides  OA  of  the  isosceles  A  OAB  (figure 
above)  is  produced  through  the  vertex  0  to  C,  so  that  OC  =  OA,  prove 
that  A  AB C  is  a  right  triangle.  What  construction  does  this  suggest  ? 

26.  The  bisectors  of  the  acute  angles  of  a  right  triangle  form  an 
angle  of  135°. 

27.  The  sum  of  the  two  exterior  angles  of  a  right 
triangle  formed  by  extending  the  hypotenuse  is  three 
right  angles. 

28.  The  hypotenuse  AB  of  a  right  tri- 
angle ABC  is  extended  in  both  directions 
so  that  BD  =  BC,  and  AE  =  A  C.   Prove 
that  Z  DCE  contains  135°. 

29.  Given  two  angles  of  a  triangle,  to 
construct  the  third  angle. 

30.  Given  a  line  /  and  two  exterior  points  A  and  B.    From  A 
and  B  drop  perpendiculars  AC  and  BD  to  I.    Connect  A  and  B. 
Prove  that  ZA  =  Z B  if  A  and  B  are  on  opposite  sides  of  1-,  and  that 
Z  A  +  Z  B  =  a  st.  Z  if  they  are  on  the  same  side  of  /. 

31.  A  sailor  often  measures  his  distance  from  a  lighthouse  by 
M  doubling  the  angle  on  the  bow."    He  notes  the  angle  which  the  line 
of  the  lighthouse  makes  with  the  course  of  the 

vessel  at  a  certain  time,  and  again  he  notes  the  time 
at  which  the  angle  is  exactly  doubled.    The  dis- 
tance the  vessel  has  traveled  in  the  interval  can 
be  determined  from  its  rate.   Show  that  this  distance  is  equal  to  the 
distance  of  the  vessel  from  the  lighthouse  at  the  time  last  taken. 
(From  the  National  Syllabus  of  Geometry.) 


ANGLE-SUM 


107 


DK 


32.  When  a  ray  of  light  strikes  a  plane  mirror  it  is  reflected  in 
accordance  with  the  following  law  :  The  angle  be- 
tween the  incident  ray  and  the  perpendicular  to  the 

mirror  is  equal  to  the  angle  between  the  reflected 

ray  and  the  same  perpendicular,  and  both  angles  lie 

in  the  same  plane.    The  first  is  called  the  angle  of 

incidence,  the  second  the  angle  of  reflection.  When 

a  ray  strikes  the  mirror  perpendicularly,  it  is  re-      ___ 

fleeted  back  over  the  same  path. 

A  ray  of  light  AB  is  incident  on  a  plane  mirror  KC  at 
right  angles  to  it.  Let  the  mirror  be  revolved  through  an 
angle  x,  taking  the  new  position  DE.  If  BH  is  _L  to  DE, 
prove  that  /.  x'  =  Z.  x,  and  that  the  angle 

between  the  incident  ray  AB  and  the         . ^ 

reflected  ray  BF is  equal  to  2  Z. x.   (This        H '""JJ^"- 

principle  is  of  importance  in  measuring          ^- — """" 

small  movements  of  rotation.)  i 

33.  A  ray  of  light  is  reflected  first  by  one  mirror  and  then  i 
by  another  in  such  a  way  that  the  path  of  the  ray  forms  an 
equilateral  triangle.    What  is  the  angle  between  the  mir- 
rors ?    (Assume  that  the  mirrors  are  perpendicular  to  the  plane  of  the 
triangle.) 

34.  A  ray  of  light  is  reflected  by  two  mirrors  successively  in  such 
a  way  that  it  returns  along  a  line  parallel  to  its  original  direction. 
What  is  the  angle  between  the  mirrors  ?  Does  this  angle  depend  upon 
the  direction  of  the  transverse  ray  from  one  mirror  to  the  other  ? 

35.  It  is  found  that  within  certain  limits 
a  ray  of  light  is  totally  reflected  from  the 
inner  surface  of  glass.    This  property  of 
glass  is  of  advantage  in  the  construction  of 

optical  instruments.    The  adjoining  figure    

represents  a  right  section  of  a  prism,  in 

the  form  of  an  isosceles  right  triangle.    A 

ray  of  light  enters  the  prism  perpendicular 

to  the  side  represented  by  one  leg  of  the 

right  triangle.    In  what  direction  does  it  leave  the  prism  ?   (Assume, 

as  is  the  case,  that  the  ray  is  within  the  limits  of  total  reflection.) 


108 


PLANE  GEOMETRY— BOOK  I 
PROPOSITION  XIX.    THEOREM 


194.   The  sum  of  the  interior  angles  of  a  polygon  is  equal  to 
as  many  straight  angles  as  the  figure  has  sides  less  two. 


Given  the  polygon  P  of  n  sides. 

To  prove  that  the  sum  of  the  interior  angles  ofP  =  (n  —  2~)  st.  A. 
Proof.    1.  Draw  all  the  diagonals  from  some  one  vertex. 
There  will  be  formed  n  —  2  triangles.  Why  ? 

2.  Now  the  sum  of  the  angles  of  each  triangle  equals  a  st.  Z. 
.'.  the  sum  of  the  angles  of  all  the  triangles  of  the  polygon 

equals  (n  -  2)  st.  A.  Why  ? 

3.  But  the  sum  of  the  angles  of  all  the  triangles  is  the  sum 
of  the  angles  of  the  polygon. 

.*.  the  sum  of  the  angles  of  the  polygon  equals  (n  —  2)  st.  A. 

EXERCISES 


1.  It  is  now  possible  to  construct  a  regular  hexagon.    Since  each 
angle    of    an   equilateral   triangle   equals   60°,   or  one   third   of   a 
straight  angle,  such  an  angle  applied 

consecutively  to  a  straight  angle  di- 
vides it  into  three  equal  parts.  If 
the  lines  of  division  are  produced 
through  the  vertex,  and  equal  dis- 
tances are  laid  off  on  the  six  rays 
from  the  vertex,  the  lines  joining  in 
succession  the  points  thus  determined 
form  a  regular  hexagon.  Prove  this.  /"  \ 

2.  Show  how  to  construct  a  regular  dodecagon ;  a  regular  24-gon. 

3.  What  regular  polygons  can  a  draftsman  draw  with  the  aid  of 
the  fixed  triangles  shown  on  page  103,  without  bisecting  angles. 


ANGLE-SUM 
PROPOSITION  XX.    THEOREM 


109 


195.  If  the  sides  of  a  polygon  are  produced  in  succession,  the 
sum  of  the  exterior  angles  thus  formed  equals  two  straight  angles. 


Given  the  polygon  Pof  n  sides,  with  its  sides  produced  in  succession. 

To  prove  that  the  sum  of  the  exterior  angles  of  P  thus  formed 
equals  two  straight  angles. 

Proof.  1.  P  has  n  vertices.  At  each  vertex  the  sum  of  the  in- 
terior angle  and  its  adjacent  exterior  angle  equals  a  straight  angle. 

2.  .  * .  the  sum  of  the  interior  and  exterior  A  of  P  =  n  st.  A. 

Why? 

3.  But  the  sum  of  the  interior  A  of  P  =  (n  —  2)  st.  A.    Why  ? 

4.  . ' .  the  sum  of  the  exterior  A  of  P  =  2  st.  A.  Ax.  3 

EXERCISES 

1.  If  the  values  of  three  angles  of  a  quadrilateral  are  70°,  80°, 
100°,  what  is  the  value  of  the  fourth  angle  ? 

2.  Three  angles  of   a  quadrilateral   are  right  angles.    Find  the 
fourth  angle. 

3.  Prove     that    if    two     opposite     AJ 

angles  of  a  quadrilateral  are  supple- 
mentary,  the   other  two    angles   are 

also  supplementary. 

4.  If  two  parallel  lines  are  cut  by 
a   transversal,   the    bisectors    of   the 
interior  angles   are  perpendicular  to 
each  other- 

5.  The  bisectors  of  the  base  angles  of  an  isosceles  triangle  and 
of  the  adjacent  exterior  angles  form  a  quadrilateral  in  which  the 
opposite  angles  are  supplementary. 


110  PLANE  GEOMETEY— BOOK  I 

6.  How  many  degrees  in  the  sum  of  the  interior  angles  of  a 
polygon  of  4  (5,  6,  7,  8,  9,  10)  sides  ? 

7.  If  these  polygons  are  regular,  how  many  degrees  in  each 
interior  angle  ?  in  each  exterior  angle  ? 

8.  In  a  regular  polygon  a  central  angle 
is  equal  to  an  exterior  angle. 

Proof,      a  +2  b  =  180°  =  x  +  2  6. 
.-.  x  =  a. 


9.  Make  a  table  giving  the  number  of  \/b 

degrees    in  the  central  angles,   the   interior 
angles,  and  the  exterior  angles  of  the  regular  polygons  known  to  you. 

10.  How  may  the  number  of  sides  of  a  regular  polygon  be  deter- 
mined, if  we  know  the  value  of  an  exterior  angle  ? 

11.  How  many  sides  has  a  polygon,  if 

(a)  The  sum  of  the  interior  angles  equals  4  rt.  A  ?  3  st.  A  ?  6  rt.  A  ? 
8  st.  A  ?   20  rt.  A  ? 

(b)  The  sum  of  the  interior  angles  is  2  (3,  4,  5,  6)  times  as  large 
as  the  sum  of  the  exterior  angles  ? 

(c)  The  sum  of  the  interior  angles  exceeds  the  sum  of  the  exterior 
angles  by  4  rt.  zt?   3  st.  A  ?   9  st.  A  ? 

(d)  The  ratio  of  each  interior  angle  to  its  adjacent  exterior  angle 
is  2:1?   3:2?   5:1?   a:&? 

(e)  Each  exterior  angle  contains  40°?  30°?  20°?  120°? 

(f)  Each  interior  angle  is  \  st.  Z?   f  st.  Z?   a  rt.  Z?   f  rt.Z? 
f  rt.  Z?    |  st.  Z? 

12.  How  does  an  increase  in  the  number  of  sides  of  a  regular 
polygon  affect  each  interior  angle  ?  each  exterior  angle  ? 

13.  Prove  the  proposition  concerning  the  sum 
of  the  angles  of  a  polygon  by  drawing  lines  from 
any  point  in  the  perimeter,  as  in  the  figure. 

14.  The  known  relation  between  a  central 
angle  of  a  regular  polygon  and  the  base  angles 
of   one   of   the   component   isosceles   triangles 

makes  it  possible  to  construct  a  regular  polygon  on  a  given  side. 
Explain. 


PARALLELOGRAMS  111 

15.  If  the  sides  of  a  polygon  are  produced  in  both  directions,  sets 
of  exterior  angles  are  formed  whose  sum  is 

8  rt.  A. 

16.  If  the  sides  of  a  polygon  are  extended 
until  they  intersect,  a  star  polygon  results.    A 
five-pointed  star  is  called  a  pentagram.    It  is 
of  historic  interest,  having  been  chosen  by  the 
followers  of  Pythagoras  as  their  badge.    Star 
polygons  may  also  be  formed  by  chords  of  cir- 
cles or  by  certain  combinations  of  polygons.    What  is  the  sum  of 
the  vertex  angles  of  a  five-pointed  star?    a  six-pointed  star?    an 
n-pointed  star? 

PARALLELOGRAMS 

196.  A  parallelogram  is  the  quadrilateral  inclosed  when  two 
pairs  of  parallel  lines  intersect  each  other. 


PRELIMINARY  PROPOSITIONS 

197.  Any   tivo    consecutive    angles   of  a  parallelogram    are 
supplementary. 

198.  The    opposite    angles    of  a   parallelogram    are    equal. 

199.  If  one  angle  of  a  parallelogram  is  a  right  angle,  the 
other  angles  are  also  right  angles. 

200.  If  two  parallelograms  have  two  adjacent  sides  and  the 
included  angle  of  one  equal  respectively  to  the  corresponding 
parts  of  the  other,  they  are  congruent. 

They  are  quadrilaterals,  and  it  can  be  readily  shown  that  they  have 
the  relation  a.  s.  a.  s.  a.  (§  144). 

Hence  a  parallelogram  may  be  constructed  when  two  sides 
and  the  included  angle  are  given. 


112  PLANE  GEOMETRY— BOOK  I 

PROPOSITION  XXI.    THEOREM 

201 .  A  diagonal  of  a  parallelogram  divides  it  into  two  con- 
gruent triangles,  and  the  opposite  sides  of  a  parallelogram  are 
equal. 


Given  the  parallelogram  ABCD,  with  the  diagonal  AC. 

To  prove   that   A  ABC  =  A  ADC,   and    that  AB=CD,   and 
AD  =  BC. 

Proof.  1.    AC  is  common  to  the  A  ABC  and  ADC. 
(To  be  completed.) 

202.  COROLLARY!.  Parallel  lines  included  between  parallel 
lines  are  equal. 

203.  COROLLARY  2.    Two  parallel  lines  intercept  equal  lengths 
on  all  lines  drawn  perpendicular  to  either  of  them. 

PROPOSITION  XXII.    THEOREM 

204.  The  diagonals  of  a  parallelogram  bisect  each  other. 


Given  the  parallelogram  ABCD,  with  the  diagonals  AC  and  BD 
intersecting  at  0. 

To  prove  that          A  O  =  OC,  and  BO  =  OD. 
Proof.   1.  A  A  OB  =  A  COD.  Why  ? 

2.  .'.  AO  =  OC,  and  BO  =  OD. 

(To  be  completed.) 


PARALLELOGRAMS 


113 


SPECIAL  PARALLELOGRAMS 

205.  A  rectangle  is  a  parallelogram  whose  angles  are  right 
angles  (see  §  199). 

206.  A  rhombus  is  an  oblique-angled  equilateral   parallel- 
ogram. 

NOTE.   A  parallelogram  with  oblique  angles  and  with  its  consecutive 
sides  unequal  is  sometimes  called  a  rhomboid. 

207.  A  square  is  an  equilateral  rectangle. 


RECTANGLE 


RHOMBUS 


SQUARE 


EXERCISES 

1.  Prove  that  the  diagonals  of  a  rectangle  are  equal. 

2.  Prove  that  if  a  parallelogram  has  equal  diagonals,  it  is  a 
rectangle. 

3.  Show  that  a  circle  can  be  circumscribed  about  any  rectangle. 

4.  Prove  that  the  diagonals  of  a  rhombus  or  a  square  are  per- 
pendicular to  each  other.   Which  parallelograms  are  also  kites? 

5.  Prove  that  if  the  diagonals  of  a  parallelogram  are  perpen- 
dicular to  each  other,  the  parallelogram  is  equilateral. 

6.  Prove  that  the   diagonals    of   a  rhombus  bisect    the  angles 
through  which  they  pass. 

7.  Prove  that  if  a  diagonal  of  a  parallelogram  bisects  the  angles 
through  which  it  passes,  the  parallelogram  is  equilateral. 

8.  Make  a  table  of  quadrilaterals  based  upon  the  relations  of 
their  diagonals. 

9.  How  does  a  carpenter  apply  Ex.  2  to  test  whether  a  window- 
casing  is  properly  "  square-cornered  "  ? 

10.  How  many  degrees  in  each  angle  of  a  rhombus  if  one  diago- 
nal is  equal  to  one  of  the  sides  ? 


114  PLANE  GEOMETRY— BOOK  I 

PROPOSITION  XXIII.    THEOREM 

208.  If  a  quadrilateral  has 

(a)  its  opposite  sides  equal,  or 

(b)  one  pair  of  opposite  sides  equal  and  parallel,  or 

(c)  its  diagonals  bisecting  each  other,  or 

(d)  its  opposite  angles  equal, 
it  is  a  parallelogram. 


(a)  Given  the  quadrilateral  ABCD,  in  which  AB  equals  CD,  and 
AD  equals  BC. 

To  prove  that  ABCD  is  a  £7. 

Proof.    1.  Draw  the  diagonal  A  C. 

2.  Then                        AABC  =  AADC.  s.  s.  s. 
For                       AB  =  CD,  and  AD  =  BC,  Hyp. 

and  AC  is  common. 

3.  .-.Zm  =  Ztt,  Why? 
and  hence                               AB  II  CD.  Why  ? 

4.  Also                             Z.p  =  Z.q,  Why? 
and  hence                               AD  II  BC.  Why  ft 

5.  .'.  ,4£C£>is  a  O.  Def. 

(b)  Given  the  quadrilateral  ABCD,  in  which  AB  equals  CD,  and 
parallel  to  CD. 

To  prove  that  ABCD  is  a  O. 

Proof.    1.  Draw  the  diagonal  A  C. 

2.  Then  A  ABC  =  A  ADC.  s.  a.  s. 

3.  .'.AD  =  BC.  Why? 
(To  be  completed.   See  (a)  above.) 


PARALLELOGRAMS  115 

(c)  Given  the  quadrilateral  ABCD,  and  the  diagonals  AC  and 
BD  bisecting  each  other  at  0. 

To  prove  that  ABCD  is  a  O. 


Proof.    1.  A  A  OB  =  A  COD.  Why  ? 

2.  .-.  Zw.=  Zw,  Why? 
and  hence  AB  II  CD.  Why  ? 

3.  Also  AB=  CD.  Why? 

4.  .-.  ABCD  is  a  E3.  Why? 

(d)  Given  the  quadrilateral  ABCD,  in  which  angle  A  equals 
angle  C,  and  angle  B  equals  angle  D. 

To  prove  that  ABCD  is  a  EH. 


Ai 


Proof.    1.  Z.A  +  Z£  +  Z  C  +  /.D  =  2  st.  A.  Why  ? 

Hyp. 


and  Z£  ==  ZJ9. 

.'./.A  +Z7?  =  ZC  +  ZD.  Ax.  2 

That  is,        Z  /I  +  Z  ZJ  =  one  half  the  sum  of  all  four  A. 
.'.  Z.A  +  Z.B  =  one  st.  Z. 

3.  .\.4Z>II£C.  Why? 

4.  In  like  manner  it  may  be  shown  that 

AB  II  CD. 

5.  .'.  ABCD  is  a  O.  Why? 


116 


PLANE  GEOMETRY— BOOK  I 


EXERCISES 

1.  Show  that  each  of  the  cases  (a),  (b),  (c),  and  (d)  of  the  above 
theorem  affords  a  method  of  constructing  a  parallelogram. 

2.  How  many  different  parallelograms  can  be  formed  by  placing 
together  two  congruent  scalene  triangles  ? 

3.  How  may  two  congruent  isosceles  triangles  be  placed  so  as  to 
form  a  rhombus  ? 

4.  Given  a  O  AJ3CD,  with  E 
its    sides   produced    in    succes- 
sion, so  that 

AE  =  CG,  and  BF  =  DH. 
Prove  that  EFGH  is  a  O. 

5.  If  in  the   above   figure  ABCD  were  a  rectangle   (rhombus, 
square),  and  AE  =  BF  =  CG  =  DH,  would  EFGH  be  a  rectangle 
(rhombus,  square)  ? 

6.  Given,  in  the  H  ABCD, 
AH  =  BE  =  CF  =  DG.    Prove 
that  EFGH  is  a  O. 

7.  If    in    the    above    figure 
ABCD  were  a  rectangle  (rhom- 
bus, square),  would  EFGH  be  a  rectangle  (rhombus,  square)  ? 

8.  What  conclusions  are  suggested  by  the  following  diagrams? 


PARALLELOGRAMS 


117 


9.  If  A  BCD  is  a  HJ,  and  MNia  any  line  through  the  intersection 
of  the  diagonals  in  the  annexed  figure,  how  many  pairs  of  congruent 
triangles  are  formed?  Give  proof.  How 
many  pairs  of  congruent  quadrilaterals? 
Give  proof.  Prove  that  the  perimeter  is 
bisected  by  such  a  line. 

10.  If  such  a  £7  as  the  above  is  cut  out 
of  cardboard  and  a  pin  is  thrust  through 

the  point  0,  the  cardboard  figure  will  balance  in  any  position.  For 
this  reason  0  is  called  the  center  of  gravity  of  the  O.  What  reason 
for  this  is  suggested  in  the  figure  ? 

11.  The  annexed  figure    shows   the 
"parallel  ruler,"  which  is  used  by  de- 
signers for  drawing  parallel  lines.    Upon 

what   principle    of    parallelograms    does    its    construction    depend  ? 

12.  Parallelogram  of  Velocities.  If  a  body  placed  at  A  has  imparted 
to  it  at  the  same  time  two  velocities,  represented  in  magnitude  and 
direction  by  the  lines  AB  and  A  C,  then  the  body  will  acquire  an  actual 
velocity  represented  by  AD,  the  diagonal  through  A  of  the  O  formed 
on  AB  and  A  Cas  consecutive  sides.    AD 

is  then  called  the  resultant  velocity,  while 
AB  and  AC  are  called  the  component 
velocities. 

For  example,  if  a  boat  is  rowed  across 
a  stream  at  the  rate  of  10  mi.  an  hour, 
the  rate  and  direction  being  represented  by  A  C,  while  the  current 
of  the  stream  is  moving  at  the  rate  of  5  mi.  an  hour,  its  rate  and 
direction  being  represented  by  AB,  then  AD  represents  the  rate  and 
the  direction  in  which  the  boat  is  actually  moving. 

Construct  with   ruler  and  protractor,  and  measure  the  missing 
elements  in  the  following  table  (referred  to  the  above  figure) : 


AB 

AC 

ZBAC 

AD 

Z.DAC 

5 

12 

90° 

p 

? 

13 

14 

60° 

9 

p 

8 

15 

45° 

p 

? 

17 

20 

60° 

p 

? 

118  PLANE  GEOMETRY—  BOOK  I 

209.  The  scheme  of  §  167  now  becomes  as  follows  : 
In  order  to  prove  two  I    ™^s  )  equal  : 

(sides  \ 
)  of  congruent  A. 
angles  / 

In  order  to  prove  two  triangles  congruent  : 

1.  Show  that  they  have  the  relation  a.  s.  a. 

2.  Show  that  they  have  the  relation  s.  a.  s. 

3.  Show  that  they  have  the  relation  s.  s.  s. 

4.  Show  that  they  have  the  relation  rt.  A  h.  1. 

5.  Show  that  they  have  the  relation  rt.  A  h.  a. 

II.  Show  that  in  a  A  they  are  ( 


HI.  Showthatthea  ,,Une, 


In  order  to  prove  that  two  lines  are  II  : 

1.  Show  that  a  transversal  makes  alt.-int.  A  equal. 

2.  Show  that  a  transversal  makes  ext.-int.  A  equal. 

3.  Show  that  a  transversal  makes  alt.-ext.  A  equal. 

4.  Show  that  a  transversal  makes  int.  A  on  the  same 

side  supplementary. 

5.  Show  that  a  transversal  makes  ext.  A  on  the  same 

side  supplementary. 

6.  Show  that  they  are  _L  to  the  same  line. 

7.  Show  that  they  are  II  to  the  same  line. 

8.  Show  that  they  are  opposite  sides  of  a  O. 

IV.  Show  that  they  are  opposite  (  S1^es  j  of  a  O. 

In  order  to  prove  that  a  quadrilateral  is  a  O  : 

1.  Show  that  its  opposite  sides  are  parallel. 

2.  Show  that  its  opposite  sides  are  equal. 

3.  Show  that  one  pair  of  opposite  sides  are  equal  and 

parallel. 

4.  Show  that  its  diagonals  bisect  each  other. 

5.  Show  that  its  opposite  angles  are  equal. 


TRANSVERSAL  THEOREM  119 

THE  TRANSVERSAL  THEOREM.   TRAPEZOIDS 
PROPOSITION  XXIV.    THEOREM 

210.  If  three  or  more  parallels  intercept  equal  parts  on  one 
transversal,  they  intercept  equal  parts  on  every  transversal. 


M 
A/ 

p 

V 

y«L>i 

V 

A     ° 

E/v     \ 

V 

A.  H 

\ 

Given  the  parallels  AB,  CD,  and  EF  intercepting  equal  parts  on 
the  transversal  MN. 

To  prove  that  they  intercept  equal  parts  on  any  other  trans- 
versal, as  PQ. 

Proof.    (Outline). 

1.  Draw  AG  and  CH  both  II  to  PQ. 

2.  Then  AG  II  CH.  §175 

3.  Now  AACG  =  ACEH.  Why? 

4.  .'.AG=CH. 

5.  But  AD  is  a  O,  and  AG  =  ED.  Why  ? 

6.  Also  CF  is  a  O,  and  CH  =  DF.  Why  ? 

7.  .'.BD  =  DF.  Why? 
In  like  manner  the  other  segments  on  PQ  are  proved  equal. 

That  is,  the  parallels  intercept  equal  parts  on  PQ. 

211.  COROLLARY.  A  line  bisecting  one  side  of  a  triangle,  and 
parallel  to  another  side,  bisects  the  third  side. 

The  third  parallel   required  by  the   above  theorem  is  constructed 
through  the  vertex  of  the  given  triangle. 


120  PLANE  GEOMETRY— BOOK  I 

PROPOSITION  XXV.    PROBLEM 

212.   To  divide  a  given  straight  line  into  any  number  of  equal 
parts. 


Given  the  straight  line  AB. 

Required  to  divide  AB  into  any  number  of  equal  parts. 
Construction.    1.  From  A  draw  the  line  AD,  making  any  con- 
venient angle  with  AB. 

2.  On  AD  lay  off  any  convenient  length  a  number  of  times 
equal  to  the  number  of  parts  into  which  AB  is  to  be  divided. 

3.  From  C,  the  last  point  thus  found  on  AD,  draw  CB. 

'4.  Through  the  other  points  of  division  on  A  D  draw  lines  par- 
allel to  CB.  These  lines  will  divide  AB  into  equal  parts.  Why  ? 

EXERCISES 

1.  A  line  AB  may  be  divided  into  equal  parts  by  the  following 
method,  which  can  be  effected  more  ^>c 
rapidly  than  the  one  preceding  :    In 

the  annexed  figure  draw  AC  II  to  BD. 
Lay  off  on  these  two  lines  equal  seg- 
ments, the  same  number  on  each. 
Join  the  corresponding  points  as 
shown.  Prove  the  construction. 

2.  A  sheet  of  ruled  paper  is  some- 
times of  use  in  dividing  a  given  line 
into  equal  parts.     Explain. 

3.  Divide  a  given  line  into  3  (4,  5,  6,  7)  equal  parts,  using  a  sheet 
of  ruled  paper. 

4.  Divide  a  given  line  in  the  ratio  of  2  :  3  (3  :  5,  1 :  4,  1 :  2  :  3). 


TKAPEZOIDS  121 

PROPOSITION  XXVI.    THEOREM 

213.  The  line  which  joins  the  mid-points  of  two  sides  of 
a  triangle  is  parallel  to  the  third  side  and  equal  to  half  the 
third  side. 


B—  G 

Given  the  triangle  ABC,  and  the  line  DE  bisecting  AB  and  AC. 
To  prove  that  DE  II  EC,  and  that  DE  =  £  BC. 

Proof.    1.  Produce  DE,  and  draw  CF  parallel  to  AB,  meet- 
ing DE  produced  in  F. 

2.  Then  AAED  =  AECF.  a.  s.  a. 
For                                     AE  =  EC,  Hyp. 

Z£  =  Zw,  Why? 

and                                        Z^  =  Z?.  Why? 

3.  *     .'.  AD  =  FC.  Why? 
But                                    AD  =  DB.  Hyp. 

.-.  FC  =  DB.  Ax.  1 

4.  Now  FC  II  DB.  Cons. 

.-.  DFCB  is  a  O.  Why? 

5.  /.  DE  II  BC,  and  D.P  =  BC.  Why  ? 

6.  But  DE  =  EF.  Why  ? 
That  is,                               DE  =  ^DF=%  BC. 

214.  Proposition  XXVI  illustrates  the  following  principle : 
In  order  to  prove  that  one  line  is  half  of  another  line, 

(1)  double  the  smaller  line,  or 

(2)  bisect  the  larger  line, 
and  proceed  according  to  §  209. 


122 


PLANE  GEOMETRY— BOOK  I 


EXERCISES 

1.  The  line  which  joins  the  mid-points  of  two  adjacent  sides  of 
any  quadrilateral  is  equal  and  parallel  to  the 

line  which  joins  the  mid-points  of  the  other 
two  sides. 

2.  What  kind  of  quadrilateral  is  formed 
by  lines  which  join  in  •  succession  the  mid- 
points of  the  sides  of  any  quadrilateral?  of 

a  kite  ?  of  a  rectangle  ?  of  a  rhombus  ?  of  a  square  ?   Prove  your 
answer  in  each  case. 

3.  The  lines  which  join  the  mid-points  of 
the  sides  of  a  triangle  divide  it  into  four 
congruent  triangles. 

4.  PROBLEM.    Given  the  angle  ABC  and 
the    point    D    within    it,    to    draw    a    line 

through  D  terminating  in  the  sides  of  the  angle  and  bisected  at  D. 

215.  A  trapezoid  is  a  quadrilateral  two  and  only  two  of 
whose  sides  are  parallel.  The  nonparallel  sides  of  a  trapezoid 
are  called  the  legs,  and  the  parallel  sides  the  bases. 


The  cross  section  of  a  canal  bank,  or  of  a  trench,  or  of  a  breakwater 
is  often  a  trapezoid.    Give  other  common  examples  of  the  trapezoid. 

216.  The  line  joining  the  mid-points  of  the  legs  is  called  the 
mid-line  of  the  trapezoid. 

217.  An  isosceles  trapezoid  is  one  whose  legs  are  equal. 

The  isosceles  trapezoid  often  arises  in  geometry  from  the  cutting  off 
of  part  of  an  isosceles  triangle  by  a  line  parallel 
to  the  base. 

218.  A  trapezium   is  a  quadrilateral 
which  has  no  two  sides  parallel. 

Have  examples  of  the  trapezium  already 
appeared  in  this  text?    Where? 


TKAPEZOIDS 

PROPOSITION  XXVII.    THEOREM 

219.    The  mid-line  of  a  trapezoid  is  parallel  to  the  bases  and 
equal  to  half  their  sum. 


V  H  C 

Given  the  trapezoid  ABCD,  and  the  line  EF  bisecting  the  legs 
AD  and  BC  in  E  and  F  respectively. 

To  prove  that  EF  II  AB  and  CD,  and  that  EF  =  ^  (AB  +  CD). 

Proof.    1.  Through  F  draw  GH  parallel  to  AD,  cutting  DC 
in  H,  and  AB  produced  in  G. 

2.  Then  ABFG  =  AHFC.  a.  s.  a. 

.'.FG  =  FH.  Why? 

3.  But  AGHD  is  a  O.  Why  ? 

.'.  4Z>  =  £#.  Why  ? 

Whence  4£  =  FG.  Ax.  5 

4.  . ' .  A  GFE  is  a  O.  Wrhy  ? 

5.  In  like  manner  EFHD  is  a  O. 

6.  ,'.EF\\AG,  and  £F  =  A  G.  Why  ? 
Also                       EF  II  Dtf ,  and  .EF  =  DH. 

7.  .'.  £F  =  J  (4G  +  £>//)  =  i-  (45  +  B<7  +  D#) 

=  ^  (,4£  +  //C  +  D/f)      Why  ? 
=  i  (45  H-  CD). 

220.  COROLLARY.    4  Ziwe  which  bisects  one  leg  of  a  trapezoid 
and  is  parallel  to  the  bases  bisects  the  other  leg  also. 


124  PLANE  GEOMETRY— BOOK  I 

EXERCISES 


1.  From  the  annexed  figures  it  appears  that : 

(a)  Any  trapezoid  may  be  broken  up  into  two  triangles  (by  a  diag- 
onal), or  into  a  rectangle  and  two  right  triangles,  or  into  a  triangle 
and  a  parallelogram.    Give  proof. 

(b)  An  isosceles  trapezoid  may  be  broken 
up  into  a  rectangle  and  two  congruent  right 
triangles.  Why  ? 


2.  Construct  a  trapezoid,  having  given  the  four  sides. 

3.  Construct  a  trapezoid,  having  given  the  two  bases  and  the  two 
base  angles. 

4.  Prove  that  each  base  of  an  isosceles  trapezoid  makes  equal 
angles  with  the  legs. 

5.  Prove  that  if  a  trapezoid  has  equal  base  angles,  it  is  isosceles. 

6.  Prove  that  the  diagonals  of  an  isosceles  trapezoid  are  equal. 

7.  Prove  that  if  a  trapezoid  has  equal  diagonals,  it  is  isosceles. 
Suggestion.    Draw  Js  from  the  ends  of  the  upper  base  upon  the  lower. 

INEQUALITIES 

221.  It  now  becomes  necessary  to  prove,  not  that  two  mag- 
nitudes are  equal,  but  that  one  of  two  magnitudes  is  greater 
than  the  other. 

The  signs  of  inequality,  >  (greater  than)  and  <  (less  than), 
are  said  to  indicate  the  sense  of  the  inequality. 

222.  Axioms   of   Inequality.     Proofs    involving    inequalities 
presuppose  the  following  axioms  (see  §  132). 

8.  The  whole  of  any  magnitude  is  greater  than  any  part  of  it. 

9.  If  equals  are  added  to  unequals,  the  results  are  unequal 
in  the  same  sense. 

Thus,  if  a  >  6,  and  c  =  d,  then  a  +  c>  6  +  d. 


INEQUALITIES  125 

10.  If  equals  are  subtracted  from  unequals,  the  remainders 
are  unequal  in  the  same  sense. 

11.  If  unequals  are  subtracted  from  equahj  the  results  are 
unequal  in  the  opposite  sense, 

That  is,  if  a  =  b  and  c  >  d,  then  a  —  c  <  b  —  d. 

12.  If  of  three  magnitudes  the  first  is  greater  than  the  second, 
and  the  second  is  greater  than  the  third,  then  the  first  is  greater 

than  the  third. 

That  is,  if  a  >  b  and  6  >  c,  then  a  >  c. 

223.  Two  inequalities  hav.e  already  been  established : 

The  exterior  angle  of  a  triangle  is  greater  than  either  remote 
interior  angle. 

In  a  right  or  an  obtuse  triangle  the  right  or  the  obtuse  angle 
is  the  greatest  angle. 

PROPOSITION  XXVIII.    THEOREM 

224.  If  two  sides  of  a  triangle  are-unequal,  the  angles  opposite 
are  unequal,  and  the  greater  angle  is  opposite  the  greater  side. 


c 

Given  the  triangle  ABC,  in  which  AB  is  greater  than  AC. 
To  prove  that  /LACB>Z.B. 

Proof.   1.  On  AB  lay  off  A  D  =  A  C. 

(This  is  possible  since  AB  >AC,  by  hyp.) 
Draw  CD. 

2.  Then  £p  =  Z<?.  Why? 

3.  But  Z.ACB  >Zp,  Ax.  8 
and                                        Z?  >Z£.  §  223 

4.  .-.  £ACB  >Z.B.  Axs.  1  and  12 


126  PLANE  GEOMETRY— BOOK  I 

PROPOSITION  XXIX.    THEOREM 

225.  If  two  angles  of  a  triangle  are  unequal,  the  sides  opposite 
are  unequal,  and  the  greater  side  is  opposite  the  greater  angle. 


C 

Given  the  triangle  ABC,  in  which  angle  B  is  greater  than  angle  C. 
To  prove  that  AC  >  A  B. 

Proof.  1.  Now   AC  =  AB,oiAC  <AB,orAC>AB. 

2.  If  AC=AB,  thenZS=ZC.  Why? 

But  this  is  contrary  to  the  hypothesis. 

3.  If  A  C  <  AB,  then  Z.B  <  Z  C.  §  224 

But  this  also  is  contrary  to  the  hypothesis. 

4.  .'.AOAB. 

(This  proposition  is  the  converse  of  Proposition  XXVIII.) 

226.  COROLLARY  1.    In  a  right  triangle  the  hypotenuse  is 
longer  than  either  of  the  legs. 

227.  COROLLARY  2.    Of  all  the  lines  that  can  be  drawn  from 
a  given  point  to  a  given  line,  the  perpendicular  is  the  shortest. 

EXERCISES 

1.  The  angles  at  the  extremities  of  the  greatest  side  of  a  triangle 
are  acute. 

2.  If  an  isosceles  triangle  is  obtuse,  the  base  is  the  longest  side. 

3.  If  the  diagonals  of  a  parallelogram  are  unequal,  the  angles  of 
the  parallelogram  must  be  oblique. 

4.  In  a  nABCD  the  side  AB>  CB.     Draw  the  diagonal  AC 
and  prove  that  Z  A  CB  >  ^A  CD. 


INEQUALITIES  127 

PROPOSITION  XXX.    THEOREM 

228.  In  any  triangle  the  sum  of  two  sides  is  greater  than 
the  third  side. 


A 

Given  the  triangle  ABC. 
To  prove  that  A  C  +  CB  >  AB. 

Proof.    1.  Produce  AC  to  D,  making  CD  ==  CB.    Draw  BD. 

2.  Then  Z^=Zy.  Why? 

3.  But  ^ABD>Z_p.  Why? 


4.  .'.  AD  >AB.  §225 

That  is,  AC  +  CB  >  AB. 

(Since  CD  =  CB.) 

229.  COROLLARY  1.  In  any  triangle  any  side  is  greater  than 
the  difference  of  the  other  two. 

Since  AC  +  CB>AB, 
.-.  AOAB-  CB.  Ax.  10 

230.  COROLLARY  2.   In  any  polygon  any  side  is  less  than  the 
sum  of  the  other  sides. 

Suggestion.  Prove  by  drawing  all  the  diagonals  from  one  end  of  the 
side  taken,  and  applying  §228  and  Ax.  12  to  the  successive  triangles 
thus  formed. 

EEMARK.  Proposition  XXX  is  often  given  among  the  self- 
evident  truths  of  geometry,  and  it  may  be  so  classified.  The 
proof  given  above  is  due  to  Euclid  (300  B.C.). 


128 


PLANE  GEOMETRY— BOOK  I 


EXERCISES 

1.  Given  a  &ABC,  with  AB>AC.    The  bisectors  of  ZJ3  and 
Z  C  meet  at  P.    Prove  that  BP  >  CP. 

2.  Given  a  &ABC,  with  AB  >  AC.  The  altitudes  from  B  and  C 
meet  at  0.    Prove  that  OB  >  OC. 

3.  The  median  to  any  side  of  a  triangle  is  less  than  half  the  sum 
of  the  other  two  sides. 

Suggestion.    Extend  the  median  its  own  length. 

4.  The  straight  line  joining  the  vertex  of  an  isosceles  triangle  to 
any  point  in  the  base  produced  is  greater  than  either  of  the  equal 
sides. 

5.  Prove    Proposition    XXX    by 
using  the  annexed  diagram,  in  which 

ED  JL  A  C.  

6.  Can    a    triangular    frame    be 

made  by  hinging  together  rods  whose  lengths  are  8  in.,  5  in.,  3  in.? 
Explain. 

7.  Given  four  rods  of  lengths  3  in.,  6  in.,  7  in.,  10  in.    How  many 
different  triangular  frames  could  be  made  by  hinging  any  three  of 
these  rods  together  at  their  extremities  ? 

8.  The  diagram   illustrates   the   mechanism    of   the   connecting 
rod  of  a  steam  engine.    BC  represents  the  connecting  rod,  CA  the 
crank,  BD  the  piston  rod.   While  the  piston  rod  BD  causes  B  to  move 
to  and  fro  along  a  straight  line,  C  de- 
scribes a  circle  with  center  A.   The  base 


AB  of  &ABC  constantly  changes.  The  extreme  positions  of  B  are 
called  the  "dead  points."  If  BC  is  4ft.  long  and  A  C  10  in.  long, 
within  what  values  does  AB  vary? 


INEQUALITIES 


129 


9.  The  straight  line  joining  the  vertex  of  an  isosceles  triangle  to  any 
point  in  the  base  is  less  than  either  of  the  equal  sides  of  the  triangle. 

10.  The  perimeter  of  a  quadrilateral  is  greater  than  the  sum  of 
its  diagonals. 

11.  If  0  is  any  point  within  a 
A  ABC,  prove  that  OB  +  OC  <  AB 
+  AC. 

(Extend  OB  to  meet  AC.) 

12.  A  ray  of  light  starting  at  A 

is  reflected  by  a  plane  mirror  m  to    ^ 
the  point  B  (see  p.  107,  Ex.  32).   Locate 
the    point    where    the    ray    strikes    the 
mirror. 

(Draw  OA  J_  to  m,  make  OA'  =  OA,  etc.) 

13.  Prove  that  the  path  of  the  ray  in 
Ex.  12,    that   is,    AC  +  CB,    is    shorter 
than  AD  +  DB,  D  being  any  other  point 
on  m. 

14.  The  diameter  of  a  circle  is  greater  than 
any  other  chord. 

(Join  the  extremities  of  the  chord  to  the  center.) 

15.  In  the  figure  AB  is  a  diameter.    Prove 
that  PA>PC,  and  that  PC>  PB. 

(Draw  OC.) 

16.  The  altitude  on  any  side  of  a  triangle  is  less  than  half  the 
sum  of  the  other  two  sides. 

17.  The  sum  of  three  altitudes  of  a  triangle  is  less  than  the 
perimeter  of  the  triangle. 

18.  Prove  Proposition  XXVIII  by  drawing  the  bisector  of  the 
/.A  to  meet  the  side  CB  in  E,  and  joining  E  to  a  point  D  on  AB 
such  that  AD  is  equal  to  A  C. 

19.  Two  regular  polygons  (§  57),  a  square  and  an  octagon,  are 
constructed  about  the  same  center  (§  95)  with  equal  radii.    Prove 
that  the  perimeter  of  the  octagon  is  greater  than  the  perimeter  of 
the  square. 


130  PLANE  GEOMETRY— BOOK  I 

PROPOSITION  XXXI.    THEOREM 

231.  If  two  triangles  have  two  sides  of  one  equal  respec- 
tively to  two  sides  of  the  other,  but  the  included  angle  of 
the  first  greater  than  the  included  angle  of  the  second,  then 
the  third  side  of  the  first  is  greater  than  the  third  side  of  the 

second. 

A  D 


F 

Given  the  triangles  ABC  and  DEF,  in  which  AB  equals  DE, 
AC  equals  DF,  and  the  angle  A  is  greater  than  the  angle  D. 

To  prove  that  BOEF. 

Proof.    1.  Construct  A  DEC  congruent  to  A  ABC,  as  shown 
in  the  figure,  and  draw  CF. 

Then  DC  falls  without  ^EDF.  Why  ? 

2.  Now  DC  =  DF.  Hyp. 

.-.  Z.DFC  =  Z.DCF.  Why? 

3.  But  Z£FC>ZZ)FC.  Why? 

.'./.EFOZ.DCF.  Ax.  1 

4.  Also  ^DCF>Z.ECF.  Why? 

.-.  AEFOZ.ECF.  Ax.  12 

5.  .'.EOEF.  §225 
That  is,                              EC  >  EF. 

Discussion.    If  F  falls  on  EC,  the  proposition  is  self-evident.    If  F  falls 
within  the  triangle  DEC,  the  proof  is  similar  to  the  one  given  above. 


INEQUALITIES  131 

PROPOSITION  XXXII.    THEOREM 

232.  If  two  triangles  have  two  sides  of  the  one  equal  respec- 
tively to  two  sides  of  the  other,  but  the  third  side  of  the  first 
greater  than  the  third  side  of  the  second,  then  the  angle  opposite 
the  third  side  of  the  first  is  greater  than  the  angle  opposite  the 
third  side  of  the  second. 

Z> 


B  E  F 

Given  the  triangles  ABC  and  DEF,  in  which  AB  equals  DE, 
AC  equals  DF,  and  BC  is  greater  than  EF. 

To  prove  that  Z  A  >  Z  D. 

Proof.    1.  Now  Z.I  =  Z.D,  or  Z.I  <  Z/>,  or  Z.I  >  Z7). 

2.  If  Z.4  =  Z/),  then  BC  =  EF.        .  Why  ? 

But  this  is  contrary  to  the  hypothesis. 

3.  If  Z.4  <  Z.D,  then  £C  <EF.^  §  231 

But  this  also  is  contrary  to  the  hypothesis. 

4.  .'./.A>Z.D. 

(This  proposition  is  the  converse  of  Proposition  XXXI.) 

EXERCISES 

1.  In  a  triangle  ABC,  AC> AB.   Equal  distances  BD  and  CE 
are  laid  off  on  BA  and  CA  respectively.    Prove  that  CD  >  BE. 

2.  In  the  A  ABC,  D  is  the  mid-point  of  the  side  BC,  and  AB  >AC. 
Prove  that  Z.ADC  is  acute. 

3.  In  the  acute  A  ABC,  AB  >  AC.  If  AD  is  the  median,  and  AE 
the  altitude  to  the  side  BC,  prove  that  E  lies  between  C  and  D. 

4.  If  any  point  P  within  the  &ABC  is  joined  to  B  and  C,  prove 


132  PLANE  GEOMETKY— BOOK  I 

PROPOSITION  XXXIII.    THEOREM 

233.  Of  two  straight  lines  drawn  from  the  same  point  in  a 
perpendicular  to  a  given  line  and  cutting  off  on  the  line'  un- 
equal segments  from  the  foot  of  the  perpendicular,  the  more 

remote  is  the  greater. 

A 


G  B  E  D 

Given  the  line  CD,  AB  perpendicular  to  CD,  and  BD  greater  than  BC. 
To  prove  that  AD>AC. 

Proof.    1.    Take  BE  =  BC}  and  draw  A  E. 

2.  Then             AE  falls  within  the  Z  BA  D,  Why? 
and                                    AABC  =  AABE.  s.  a.  s. 

3.  .                   .\AE  =  AC.  Why? 

4.  Now                   Z AEB  is  an  acute  Z.  Why? 

.'.  /-AED  is  an  obtuse  Z.  Why  ? 

Also                          Z.ADB  is  an  acute  Z.  Why  ? 

5.  .'.AD>AE.  §225 
That  is,                             AD>AC.  Ax.  1 

234.  COROLLARY.    Only  two  equal  straight  lines  can  be  drawn 
from  a  point  to  a  straight  line  ;  and  of  two  unequal  lines  from 
a  point  to  a  line,  the  greater  cuts  off  the  greater  segment  from  the 
foot  of  the  perpendicular. 

Prove  from  Proposition  XXXIII  by  the  Indirect  Method. 

235.  The  distance  between  two  points  is  the  length  of  the 
line-segment  joining  them.  §  228 

236.  The  distance  from  a  point  to  a  line  is  the  length  of  the 
perpendicular  let  fall  from  that  point  upon  the  line.          §  227 


INEQUALITIES 


133 


237.  The  distance  between  two  parallel  lines  is  the  length  of  the 
segment  of  a  common  perpendicular  cut  off  between  the  two  lines. 

238.  The  altitude  of  a  paral- 
lelogram or  of  a  trapezoid  is 
the    perpendicular    distance 
between  the  bases,  as  MN. 
(AD  and  EC  are  called  the 
lower   and   the   upper   base 
respectively.) 

PQ  is  also  an  altitude  of  the  OABCD,  the  corresponding  bases  being 
AB  and  CD. 

EXERCISES 

1.  In  a  quadrilateral  DEFH,  DH  =  EF,  and  ZH>ZF.    Prove 
that  DF  >  EH. 

2.  In  a  quadrilateral  DEFH,  DH  =  EF,  and  DF  >  EH.    Prove 
that  ZH>ZF. 

3.  If  P  is  any  point  within  the  A  ABC,  prove  that  PA  +  PB  +  PC 
is  less  than  the  perimeter,  but  greater  than  half  the  perimeter,  of 
the  triangle  (see  Ex.  11,  p.  129). 

4.  If  P  is  any  point  within  a  quadrilateral,  prove  that  the  sum  of 
its  distances  from  the  vertices  is  not  less  than  the  sum  of  the  diag- 
onals.   When  is  it  equal  to  the  sum  of  the 

diagonals  ? 

5.  The   sum  of   the  distances  of  any 
point  within  a  polygon  from  the  vertices 
is  greater  than  half  the  perimeter  of  the 
polygon. 

6.  Two  plane  mirrors  m  and  n  meet  at 
an  angle.    The  points  A  and  B  are  within 
the  angle  mn.    The  figure  shows  the  path 

of  a  ray  of  light  emanating  from  A  and  reflected  by  the  mirrors  m 
and  n  through  B.  Take  any  other  two  points,  one  each  in  m  and  n, 
join  them  by  a  line-segment,  and  draw  the  lines  from  the  point  in  m 
to  A  and  from  the  point  in  n  to  B.  Show  that  the  length  of  this 
path  from  A  to  B  is  greater  than  the  path  of  the  ray  of  light.  This 
exercise  illustrates  the  fact  that  light  travels  by  the  shortest  path. 


A' 


A 


134  PLANE  GEOMETRY— BOOK  I 

COLLINEARITY  AND  CONCURRENCE 
PROPOSITION  XXXIV.    THEOREM 

239.  All  points  in  the  perpendicular  bisector  of  a  line  are 
equidistant  from  the  extremities  of  the  line ;  and  conversely, 
all  points  equidistant  from  the  extremities  of  the  line  lie  in  the 

perpendicular  bisector. 

D 

/\ 
/  A 


m 


\ 


A  G  B 

(a)  Given  DC,  the  perpendicular  bisector  of  AB. 

To  prove  that  all  points  in  DC  are  equidistant  from  A  and  B. 
Proof.    1.   Take  any  point  in  DC,  as  E.    Draw  AE  and  BE. 

2.  Then  AAEC  =  ABEC.  Why? 

3.  .'.AE  =  BE.  Why? 

4.  .'.  since  E  is  any  point  in  DC,  all  points  in  DC  are  equi- 
distant from  A  and  B. 

(b)  Given  the  line  AB. 

To  prove  that  all  points  equidistant  from  A  and  B  lie  in  the 
perpendicular  bisector  of  AB. 

Proof.    1.  Take  any  such  equidistant  point  as  F.    Draw  FA 
and  FB.    Then  FA  =  FB.   Draw  FC  bisecting  AB. 

2.  Then  AAFC  =  ABFC.  Why? 

3.  .-.ZW  =  ZTZ.  Why? 

.'.  FCA.AB.  Why? 

4.  That  is,  the  point  F  lies  in  the  perpendicular  bisector  of  AB. 

5.  But  F  is  any  point  equidistant  from  A  and  B. 

.'.  all  such  points  lie  in  the  perpendicular  bisector  of  AB. 


COLLINEARITY  AND  CONCURRENCE         135 

PROPOSITION  XXXV.    THEOREM 

240.  All  points  in  the  bisector  of  an  angle  are  equidistant 
from  the  sides  of  the  angle ;  and  conversely,  all  points  equidis- 
tant from  the  sides  of  an  angle  lie  in  the  bisector  of  the  angle. 


(a)  Given  the  angle  AOB  bisected  by  OC. 

To  prove  that  all  points   in   OC   are   equidistant  from   OA 
and   OB. 

Proof.    1.  Take  any  point  in  OC,  as  M.    Draw  JJ/P_Lto  OA, 
and  M Q  _L  to  OB. 

2.  Then  AMOP  =  AMOQ.  Why? 

3.  .'.MP  =  MQ.  Why? 

4.  .'.  since  M  is  any  point  in  OC,  all  points  in  OC  are  equi- 
distant from  OA  and  OB. 

(b)  Given  the  angle  A  OB. 

To  prove  that  all  points  equidistant  from  OA  and  OB  lie  in 
the  bisector  of  the  Z  A  OB. 

Proof.    1.  Take   any  such   equidistant   point   as   N.    Draw 
NR  _L  to  OA,  and  NS  _L  to  OB.    Then  NR  =  NS.    Draw  NO. 

2.  Then  ANOR  =  ANOS.  Why? 

3.  .'.  Z  h  =  Z  k,  that  is,  NO  bisects  Z.AOB. 

4.  That  is,  the  point  N  lies  in  the  bisector  of  the  Z.AOB. 

5.  But  N  is  any  point  equidistant  from  the  sides  of  the  /-  A  OB. 
.'.all  points  equidistant  from  OA  and  OB  lie  in  the  bisector 

of  the  /.AOB. 


136  PLANE  GEOMETRY— BOOK  I 

PROPOSITION  XXXVI.    THEOREM 

241.   The  perpendicular  bisectors  of  the  sides  of  a  triangle 
meet  in  a  point  equidistant  from  the  three  vertices. 


C  E  & 

Given  the  triangle  ABC,  with  DH,  El,  and  FG  the  perpendicular 
bisectors  of  the  sides  AB,  BC,  and  CA  respectively. 

To  prove  that  DH,  El,  and  FG  meet  in  a  point  equidistant 
from  A,  B,  and  C. 

Proof.    1.  DH  and  FG  will  intersect.  §  178 

(Since  they  cannot  be  II,  being  J_  to  two  intersecting  lines.) 
Call  the  point  of  intersection  0. 

2.  Then          0  is  equidistant  from  A  and  B.  §  239  (a) 
Also                 0  is  equidistant  from  A  and  C.  §  239  (a) 

3.  .'.  0  is  equidistant  from  B  and  C.  Ax.  1 

.'.  Olies  in  EL  §  239  (b) 

That  is,  DH,  El,  and  FG  meet  in  0,  which  is  equidistant  from 
A,  B,  and  C. 

242.  The  point  0  is  called  the  circumcenter  of  the  A  ABC. 

243.  Three  or  more  lines  which  meet  in  a  point  are  said  to 
be  concurrent. 

Thus,  from  the  above  proposition,  the  perpendicular  bisectors  of  the 
sides  of  a  triangle  are  concurrent. 

244.  Three  or  more  points  which  lie  in  the  same  line  are 
said  to  be  collinear. 

For  example,  from  §  239,  points  equidistant  from  the  extremities  of  a 
line  are  collinear. 


COLLINEAKITY  AND  CONCUKRENCE         137 

PROPOSITION  XXXVII.    THEOREM 

245.   The  bisectors  of  the  angles  of  a  triangle  are  concurrent 
in  a  point  equidistant  from  the  three  sides. 

A 


B 


Given  the  triangle  ABC,  with  AD,  BE,  and  CF  the  bisectors  of 
the  angles  A,  B,  and  C  respectively. 

To  prove  that  AD,  BE,  and  CF  are  concurrent  in  a  point  equi- 
distant from  the  sides  of  the  triangle. 

Proof.    1.  AD  and  BE  will  intersect. 

(Since  they  cannot  be  ||,  the  sum  of  the  A  which  they  make  with  the 
transversal  AB  being  less  than  two  rt.  A.) 

Call  the  point  of  intersection  0. 

2.  Then       0  is  equidistant  from  AC  and  AB.          §  240  (a) 
Also  0  is  equidistant  from  AB  and  BC.          §  240  (a) 

3.  .'.  0  is  equidistant  from  AC  and  BC.  Ax.  1 

.'.  Olies  in  CF.  §  240  (b) 

That  is,  AD,  BE,  and  CF  are  concurrent  in  0,  which  is  equi- 
distant from  the  three  sides. 

246.  The  point  0  is  called  the  incenter  of  the  A  ABC. 

EXERCISES 

1.  What  construction  is  suggested  by  Proposition  XXXVI  ? 

2.  Does  the  incenter  of  a  triangle  always  fall  within  the  triangle  ? 

3.  In  the  l\ABC  show  that  the  bisectors  of  the  interior  angle  A 
and  the  exterior  angles  at  B  and  C  are  concurrent  in  a  point  equi- 
distant from  the  sides  of  the  triangle.  This  point  is  called  an  excenter 
of  the  &ABC.    How  many  excenters  has  a  triangle? 


138  PLANE  GEOMETRY— BOOK  I 

PROPOSITION  XXXVIII.    THEOREM 

247.  The  medians  of  a  triangle  are  concurrent  in  a  point 
which  is  two  thirds  of  the  distance  from  each  vertex  to  the 
middle  of  the  opposite  side. 

A 


C  D  B 


Given  the  triangle  ABC,  with  the  medians  AD,  BE,  and  CF  to  the  sides 
BC,  CA,  and  AB  respectively. 

To  prove  that  AD,  BE,  and  CF  are  concurrent  in  a  point  two  thirds  of 
the  distance  from  each  vertex  to  the  middle  of  the  opposite  side. 

Proof.    1.  AD  and  BE  will  intersect. 

(Since  one  of  them  joins  two  points  which  lie  on  opposite  sides  of  the  other.) 
Name  the  point  of  intersection  O.   Bisect  A  0  and  BO  in  G  and 
H  respectively.   Draw  ED  and  GH,  also  EG  and  DH. 

2.  Then  DE  II  AB,  and  DE  =  ±  AB.  Why? 
Also                         GH  II  AB,  and  GH  =  1  AB.  Why  ? 

.'.DE  =  GH  (Ax.  1),  and  DE  II  GH.  Why  ? 

.'.EGHDis&O.  Why? 

Whence  EO  =  OH  =  1  OB.  Why  ? 

3.  .'.  BO  =  20E  =  %BE. 
Also  AO  =  %AD. 

That  is,  AD  and  BE  intersect  in  a  point  which  is  two  thirds  of 
the  distance  froni  each  vertex  to  the  middle  of  the  opposite  side. 

4.  In  like  manner  it  can  be  proved  that  CF  and  AD  intersect  in 
a  point  which  is  two  thirds  of  the  distance  from  each  vertex  to  the 
middle  of  the  opposite  side.    That  is,  CF  passes  through  O. 

.'.  AD,  BE,  and  CF  are  concurrent  in  a  point  two  thirds  of  the 
distance  from  each  vertex  to  the  middle  of  the  opposite  side. 

248.   The  point  0  is  called  the  centroid  of  the  &ABC. 


COLLINEARITY  AND  CONCUKBENCE          139 

PROPOSITION  XXXIX.    THEOREM 

249.    The  altitudes  of  a  triangle  are  concurrent. 

W 


Given  the  triangle  ABC,  with  the  altitudes  AL,  BM,  CN  respectively. 
To  prove  that  AL,  BM,  and  CN  are  concurrent. 

Proof.    1.  Through  A,  B,  and  C  respectively  draw  EF,  FD,  DE  II  to 
BC,  CA,  and  AB  respectively. 

2.  Then  FBCA  is  a  a.  Why  ? 

.'.FB  =  AC.  Why? 

Also                                    BDCA  is  a  a.  Why  ? 

.\BD  =  AC.  Why? 

3.  .*.  FB  =  BD.  Ax.  1 

4.  Now  BM±AC.  Hyp. 

.\BM±FD.  Why? 

That  is,        BM  is  the  perpendicular  bisector  of  FD. 

(To  be  completed.) 
250.  The  point  0  is  called  the  orthocenter  of  the  A  ABC. 

EXERCISES 

1.  Cut  a  triangle  out  of  cardboard.    Find  its  centroid  and  thrust 
a  pin  through  this  point.    The  triangle  will  balance  in  whatever 
vertical  position  it  is  placed.    This  shows  that  the  centroid  is  the 
center  of  gravity  of  the  triangle.    A  reason  why  this  should  be  so 
will  be  given  in  Book  IV. 

2.  In  an  equilateral  triangle  the  incenter,  the  circumcenter,  the 
orthocenter,  and  the  centroid  are  the  same  point. 


140 


PLANE  GEOMETRY— BOOK  I 


FIG.  1 


MISCELLANEOUS  EXERCISES 

1.  The  perpendiculars  upon  the  legs  of   an  isosceles  triangle 
from  the  mid-point  of  the  base  are  equal. 

2.  The  sum  of  the  perpendiculars  from  a  point  in  the  base  of 
an  isosceles  triangle  to  the  legs  is  equal  to  the  alti- 
tude upon  one  of  the  legs  (Fig.  1). 

3.  The  sum  of  the  perpendiculars  from  any  point 
within  an  equilateral  triangle  to  the  three  sides  is 
equal  to  the  altitude  of  the  triangle  (Fig.  2). 

4.  If  two  medians  of  a  triangle  are  equal,  the 
triangle  is  isosceles. 

5.  The    perpendiculars    from  the   mid-points   of 
two  sides  of  a  triangle  upon  the  third  side  are 
equal. 

6.  The  lines  drawn  through  two  opposite 
vertices  of  a  parallelogram  to  the  mid-points 
of  the  opposite  sides  divide  one  of  the  diago- 
nals into  three  equal  parts  (Fig.  3). 

7.  If  upon  the  three  sides  of  a  triangle 
equilateral  triangles  are  constructed 

lying  outside  the  given  triangle,  the 
lines  joining  the  outer  vertices  of 
the  equilateral  triangles  to  the  op- 
posite vertices  of  the  given  triangle 
are  equal. 

8.  If  on  one  leg  of  an  isosceles 

triangle,  and  on  the  other  leg  produced,  equal 
lengths  are  laid  off  on  opposite  sides  of  the 
base,  the  line  joining  the  points  thus  deter- 
mined is  bisected  by  the  base  (Fig.  4). 

9.  The  line  joining  the   mid-points  of  the 
diagonals  of  a  trapezoid  is  equal  to  half  the 

difference  of  the  bases. 

FIG.  4 
10.  Two  triangles  are  congruent  if  they  have 

a  side  and  the  altitude  and  the  median  on  that  side  in  one  equal  to 
corresponding  parts  in  the  other. 


FIG.  2 


FIG.  3 


MISCELLANEOUS  EXERCISES 


141 


11.  Composite  Diagrams.  Examine  each  of  the  following  figures 
with  reference  to  (a)  equal  lines,  (b)  equal  angles,  (c)  congruent 
triangles,  (d)  parallel  lines,  (e)  parallelograms, 
(f)  particular  lines  (bisectors,  medians,  alti- 
tudes, etc.),  (g)  symmetrical  figures,  (h)  regu- 
lar figures. 


FIG.  9 


Fig.  5.    ABC  and  ABD  are  congruent  isosceles  triangles.    AB  is 
divided  into  four  equal  parts. 

Fig.  6.    A  BCD  is  a  square. 

AE  =  AM  =  BF  =  BG  =  CH=CI  =  DK  =  DL. 

Fig.  7.    A  BCD  is  a  square.    ABF  and  CDE  are  equilateral  tri- 
angles. 

Fig.  8.    ABCDEF  is  a  regular  hexagon. 

Fig.  9.    A  BCD  is  a  square,  with  equilateral  triangles  on  its  sides. 


142  PLANE  GEOMETRY— BOOK  I 

12.  The  bisectors  of  two  exterior  angles  of  a  triangle  meet  at 
an  angle  which  is  equal  to  one  half  the  third  exterior  angle. 

13.  Two  mirrors,  ml  and  m2, 
are  set  so  as  to  form  an  acute 
angle   with  each  other.    A  ray 
of  light  is  reflected  by  m1  so  as 
to  strike  'mr    The  ray  is  again 
reflected  by  m2  and  crosses  its 
first  path.    Prove  that  Z  x  —  2  Za. 
(This  is  the  principle  underlying 
several  important  optical  instru- 
ments, such  as  the  sextant.) 

14.  The  circumcenter  of  a  right  triangle  is  the  mid-point  of  the 
hypotenuse ;  of  an  obtuse  triangle,  a  point  without  the  triangle. 

15.  If  the  bisectors  of  the  angles  of  a  quadrilateral  are  concurrent, 
the  sum  of  two  opposite  sides  equals  the  sum  of  the  other  two  oppo- 
site sides. 

16.  Two  trapezoids  are  congruent  if  their  sides  are  respectively 
equal. 

17.  The  lines  which  join  in  succession  the  mid-points  of  the  sides 
of  an  isosceles  trapezoid  inclose  a  rhombus. 

18.  If  the  bisectors  of  two  opposite  angles  of  a  quadrilateral  form  a 
straight  line,  the  bisectors  of  the  other  two  angles  meet  on  that  line. 

19.  If  the  bisectors  of  the  interior  angles  of  a  quadrilateral  are  not 
concurrent,  they  inclose  a  quadrilateral  of  which  the  opposite  angles 
are  supplementary. 

20.  Prepare  a  summary  of  the  most  important  topics  in  Book  I. 

21.  Give  a  list  of  construction  lines  that  have  been  helpful  in 
Book  I. 

22.  What  methods  of  proving  lines  or  angles  equal  have  been 
given  so  far? 

23.  Prepare  a  list  of  all  propositions  which,  directly  or  indirectly, 
justify  Proposition  XXXVIII.    (These  may  be  arranged  by  number, 
according  to  their  relative  dependence,  in  the  form  of  a  pyramid.) 

24.  Prepare  a  summary  of  the  applied  problems  given  in  the  exer- 
cises of  Book  I. 


BOOK  II 

THE  CIRCLE 

251.  Preliminary  Propositions.  The  following  properties  of 
circles  have  been  either  proved  or  taken  for  granted  previously : 

1.  A  circle  can  be  drawn  about  any  given  point  as  a  center, 
with  a  radius  equal  to  any  given  line. 

2.  Radii  of  the  same  circle  are  equal  (§  60). 

3.  Circles  which  have  equal  radii  are  equal  (§  61). 

4.  A  point  is  within,  on,  or  without  a  circle,  according  as  the 
line  which  joins  the  point  to  the  center  is  less  than,  equal  to,  or 
greater  than  the  radius  of  the  circle  (§  62). 

5.  A  secant  to  a  circle  intersects  the  circle  in  two  and  only 
two  points  (§68). 

6.  If  the  center  segment  of  two  circles  is  less  than  the  sum 
but  greater  than  the  difference  of  their  radii,  the  circles  intersect 
in  two  and  only  two  points  (§  71). 

7.  In  the  same  circle  or  in  equal  circles  : 

Equal  central  angles  intercept  equal  arcs  (§  85). 

Equal  arcs  are  intercepted  by  equal  central  angles  (§  85). 

Equal  chords  subtend  equal  arcs  (§  86). 

Equal  arcs  are  subtended  by  equal  chords  (§  86). 

8.  A  central  angle  is  measured  by  its  intercepted  arc  (§  94). 

9.  In  the  same  circle,  or  in  equal  circles,  the  greater  central 
angle  intercepts  the  greater  arc,  and  the  greater  arc  is  inter- 
cepted by  the  greater  central  angle  (§  94). 

10.  A  diameter  of  a  circle  bisects  the  circle  (§  63). 

NOTE.  A  circle  may  be  conveniently  named  by  its  center.  Thus  a 
circle  whose  center  is  0  will  hereafter  be  called  "the  circle  O."  The 
word  "circle"  is  often  replaced  by  the  symbol  O. 

143 


144 


PLANE  GEOMETKY— BOOK  II 


EXERCISES 

1.  Two  points  P  and  Q  are  2  cm.  apart.   Draw  a  circle,  of  radius 
3  cm.,  passing  through  P  and  Q. 

2.  The   centers  of  two   circles  are  4  cm. 
apart.    The  radii  of  the  circles  are  3  cm.  and 
5  cm.  respectively.    Describe  the  relative  posi- 
tion of  the  two  circles. 


3.  If,    in    Fig.  1,    Z  x  =  Z.  y,    prove    that 
arc  ABC  =  arc  BCD. 

4.  If,  'in  Fig.  2,  chord  AB  =  chord  BC  = 
chord  CD,  prove  that  chord  AC  =  chord  ED. 

5.  In  the  same  figure,  prove  that  /.A  =  ZD. 
What  other  angles  are  equal  ? 

6.  If  the  sides  of  an  inscribed  polygon  are 
equal,  prove  that  the  polygon  is  equiangular 
(see  Ex.  5). 

7.  InFig.3,JBC  =  ^lZ>/ProvethatylJB=CZ). 

8.  If-  two  equal  chords  of  a  circle  intersect, 
the  segments  of  one  are  equal  respectively  to 
the  segments  of  the  other. 

9.  In  Fig.  ±,AB  =  CD.  Prove  that  A  ABD= 
A  CDB,  and  that  A  A  OD  =  A  COB. 

10.  In  the  same  figure  prove  that  &BOD  is 
isosceles. 


FIG.  1 


FIG.  2 


FIG. 


11.  If  from  a  point  0  on  a  circle  (Fig.  5) 
a  chord  OA  and  a  diameter  OB  are  drawn, 
prove  that  a  radius  parallel  to  OA  bisects  the  arc  AB.  (Draw  PA.) 


FIG.  4 


FIG.  5 


BOOK  II 


145 


CHORDS  AND  SECANTS 
PROPOSITION  I.    THEOREM 

252.  A  diameter  perpendicular  to  a  chord  bisects  the  chord 
arid  the  arcs  subtended  by  it. 

D 


Given  the  circle  0,  with  a  diameter  DC  perpendicular  to  the 
chord  AB  at  M,  cutting  the  minor  arc  ACB  at  C,  and  the  major  arc 
ADB  at  D. 

To  prove  that  AM  =  BM,  arc  AC  =  arc  CB,  and  arc  AD  = 
arc  DB. 

Draw  the  radii  OA  and  OB. 

rt.  A  A  OM  =  rt.  A  BOM.  Why  ? 

.'.AM  =  BM.  Why? 

Z.AOM  =  S/.BOM,  Why? 

,:.  Z.AOD  =  /.BOD.  Why? 

5.          .'.  arc  vlC  =  arc  CB,  and  arc  AD  =  arc  DB.  §  85 

253.  COROLLARY  1.  A  diameter  which  bisects  a  chord  is  per- 
pendicular to  it. 

254.  COROLLARY  2.   The  perpendicular  bisector  of  a  chord 
passes  through  the  center  of  the  circle. 

Discussion.    Show  how  to  bisect  an  arc  whose  center  is  not  given. 

255.  A  circle  is  said  to  be  circumscribed  about  a  polygon  when 
the  sides  of  the  polygon  are  chords  of  the  circle.   The  polygon 
is  said  to  be  inscribed  in  the  circle. 


Proof.    1. 
2.  Then 
3. 

4.  Also 
and 


146       PLANE  GEOMETRY— BOOK  II 

PROPOSITION  II.    PROBLEM 
256.   To  circumscribe  a  circle  about  a  given  triangle. 


Given  the  triangle  ABC. 

Required  to  circumscribe  a  O  about  the  A  ABC. 

Construction.  1.  Construct  the  perpendicular  bisectors  of  two 
of  the  sides  and  produce  them  until  they  meet  in  0  (§  241). 

2.  About  0,  with  a  radius  equal  to  OB,  describe  a  O. 

The  O  will  pass  through  A,  B,  and  C,  since  0  is  equidistant 
from  A,  B,  and  C.  (Why?)  It  is  therefore  circumscribed  about 
'the  A  ABC,  by  the  definition  of  a  circumscribed  circle.  §  255 

(Under  what  condition  does  the  circumcenter  lie  on  a  side  of  the  tri- 
angle ?  without  the  triangle  ?) 

257.  COROLLARY  1.    Only  one  circle  may  be  circumscribed 
about   a  given  triangle. 

For,  if  there  were  two  such  circles,  they  would  intersect  in  three 
points,  which  is  impossible  (§  71). 

258.  COROLLARY  2.   One  and  only  one  circle  may  be  drawn 
through  any  three  points  not  in  the  same  straight  line. 

259.  COROLLARY  3.  A  circle  cannot  be  passed  through  three 
points  in  the  same  straight  line.  §  68 

260.  COROLLARY  4.   The  center  of  a  given  arc  may  be  found 
by  drawing  two  chords  and  finding  the  point  of  intersection  of 
their  perpendicular  bisectors. 

261 .  A  triangle  is  said  to  be  inscribed  in  a  semicircle  when  one 
of  its  sides  is  the  diameter  of  the  semicircle  and  the  opposite 
vertex  lies  on  the  semicircle. 


CHOBDS  AND  SECANTS 


147 


262.  COROLLARY  5.  A  triangle  inscribed  in  a  semicircle  is  a 
right    triangle,    the  diameter  being 
the  hypotenuse. 


Proof.     OA  =  OB  =  OC. 

.-.  Z  a  =  Z  c,  and  Z  6  =  Z  d. 


Why  ? 
Why  ? 
Why? 


That  is,  the  triangle  is  a  right  triangle,  and  the  diameter  is  its  hypotenuse. 

263.  COROLLARY  6.    PROBLEM.   At  a  given  point  in  a  given 
line  to  erect  a  perpendicular  to  that  line.  ^ 

(Second  construction,  see  §  158.) 

Given  the  line  AB  and  the  point  C  in  it. 
Kequired  to  erect  a  JL  to  AB  at  C.  Take  any 
convenient  point  D  not  on  AB,  and  with  a 
radius  DC  describe  a  O  cutting  AB  in  E. 

Draw  the  diameter  through  E,  and  let  .F  be  its    A-    -^  N^«. ... 

other  extremity.    Draw  CF.   Then  CF  _L  CE. 

For  A  FCE  is  a  right  A  with  EF  as  its  hypotenuse.    (Why  ?) 

EXERCISES 

1.  The  perpendicular  bisectors  of  the  sides 
of  an  inscribed  polygon  are  concurrent. 

2.  If,  in  the  diagram,  the  two  circles  are 
concentric,  prove  that  AB  =  CD. 

3.  By  means  of  a  circular  object  (e.g.  a  coin) 
draw  an  arc  of  a  circle.    Locate  its  center. 

4.  A  circle  may  be  circumscribed  about 
an  isosceles  trapezoid. 

5.  Two  chords  perpendicular  to  a  third 
chord  at  its  extremities  are  equal. 

6.  The  radius  of  a  circle  circumscribed 
about  an  equilateral  triangle  is  equal  to 
two  thirds  the  altitude. 

7.  If  two  plane  mirrors  are  inclined  at 
an  angle  and  an  object  is  placed  within  the 
angle,  several  images  of-  the  object  will 
appear.  These  will  lie  on  a  circle.  Explain. 


148  PLANE  GEOMETRY— BOOK  II 

PROPOSITION  III.    THEOREM 

264.  In  the  same  circle,  or  in  equal  circles,  equal  chords  are 
equally  distant  from  the  center;  and,  conversely,  chords  equally 
distant  from  the  center  are  equal. 

C 


Given  the  circle  0  in  which  the  chords  AB  and  CD  are  equal. 
To  prove  that  A B  and  CD  are  equally  distant  from  the  center. 

Proof.    1.  Construct  OE  _L  to  AB,  OF  J_  to  CD,  and  draw 

the  radii  OA  and  OD. 

(To  be  completed.) 

Conversely,  given  the  circle  0,  with  AB  and  CD  equally  distant 
from  the  center  0. 

To  prove  that  AB  =  CD. 

Proof.  (To  be  completed.) 

EXERCISES 

1.  If  the  sides  of  an  inscribed  polygon  are  equally  distant  from 
the  center,  the  polygon  is  equilateral. 

2.  Can  a  circle  be  circumscribed  about  any  given  parallelogram  ? 
Can  a  parallelogram  be  inscribed  in  a  circle  ?   Explain  your  answer. 

3.  If  two  intersecting  chords  make  equal  angles  with  the  diameter 
passing  through  the  point  of  intersection,  the  two  chords  are  equal. 

4.  A  circle  whose  center  is  on  the  bisector  of  an  angle  cuts  equal 
chords,  if  any,  from  the  sides  of  the  angle. 

5.  Two  radii,  OB  and  OC,  of  a  circle  intercept  a  minor  arc. 
A  point  D  in  the  arc  is  joined  to  the  mid-points  E  and  F  of  the 
radii.    If  DE  =  DF,  prove  that  arc  CD  =  arc  DB. 


CHORDS  AND  SECANTS  149 

PROPOSITION  IV.    THEOREM 

265.  In  the  same  circle,  or  in  equal  circles,  the  greater  of  two 
minor  arcs  is  subtended  by  the  greater  chord  ;  and  conversely, 
the  greater  chord  subtends  the  greater  minor  arc. 


Given,  in  the  circle  O,  the  arc  AB  greater  than  the  arc  A'Br. 
To  prove  th a t  chord  AB>  chord  A  'B'. 

Proof.    1.       Draw  the  radii  OA,  OB,  OA',  OB'. 

2.  In  the  A  A  OB  and  A' OB', 

AO  =  A'0,  and  BO  =  B'O.  Why? 

Also  Zra>Zra'. 

(Since  arc  AB  >  arc  A'B',  by  Hyp.) 

3.  .\AB>A'B'.  §231 

Conversely,  given,  in  the  circle  O,  the  chord  AB  greater  than  the 
chord  A'B'. 

To  prove  that  arc  AB>  arc  A 'B'. 

Proof.    1.       Draw  the  radii  OA,  OB,  OA',  OB'. 

2.  In  the  A  A  OB  and  A' OB', 

AO  =  A '0,  and  BO  =  B'O.  Why  ? 

Also                                    AB>A'B'.  Hyp. 

3.  .'.Z.AOB>^A'OB'.  §232 
.-.  arc  AB > arc  A'B'.  Why  ? 


150       PLANE  GEOMETRY— BOOK  II 

PROPOSITION  V.    THEOREM 

\ 

266.  In  the  same  circle,  or  in  equal  circles,  if  tivo  chords  are 
unequal,  they  are  unequally  distant  from  the  center,  and  the 
greater  chord  is  at  the  less  distance. 


Given  the  chords  AB  and  CD  in  the  circle  0,  AB  being  greater 
than  CD ;  given  also  OF  perpendicular  to  AB,  and  OH  perpendicular 
to  CD. 

To  prove  that  OF  <  OH. 

Proof.    1.          Minor  arc  CD  <  minor  arc  AB.  §  265 

2.  Take  the  point  E  on  the  minor  arc  AB,  so  that 

arc  A  E  =  arc  CD. 

Draft  AE. 
Draw  OK  _L  toAE. 

Then  chord  AE  =  chord  CD,  Why  ? 

and  hence  OK  =  OH.  Why  ? 

3.  Now  all  points  of  the  chord  AE  lie  upon  the  opposite 
side  of  the  chord  AB  from  the  center  O. 

.'.  OK  must  intersect  AB  in  some  point,  as  L. 

4.  .'.OL<OK.  Why? 

5.  But  OF<OL.  Why? 

.'.OF<OK,  Why? 

or  OF<  OH. 


CHOKDS  AND  SECANTS  151 

PROPOSITION  VI.    THEOREM 

267.  In  the  same  circle,  or  in  equal  circles,  if  two  chords  are 
unequally  distant  from  the  center,  they  are  unequal,  and  the 
chord  at  the  less  distance  is  the  greater. 


Given  the  chords  AB  and  CD  in  a  circle  0 ;  also  OF,  the  per- 
pendicular to  AB,  less  than  OH,  the  perpendicular  to  CD. 

To  prove  that  .  AB>  CD. 

Proof.  1.  Now  AB  =  CD,  or  <  CD,  or  >  CD. 

2.  If  AB=  CD,  then  OF=  OH.  Why  ? 

But  this  is  contrary  to  the  hypothesis. 

3.  If  AB  <  CD,  then  OF>  OH.  §  266 

But  this  also  is  contrary  to  the  hypothesis. 

4.  .'.AB>CD. 

EXERCISES 

1.  The  shortest  chord  that  can  be  drawn  through  a  point  within 
a  circle  is  perpendicular  to  the  diameter  at  that  point. 

2.  The  diameter  is  the  greatest  chord  of  a  circle. 

3.  If  a  chord  is  drawn  parallel  to  a  diameter,  the  arcs  intercepted 
between  the  chord  and  the  diameter  are  equal.    (Draw  radii  to  the 
extremities  of  the  chord.) 

4.  Two  parallel  chords  intercept  equal  arcs  on  a  circle  (see  Ex.  3). 

5.  If  an  arc  is  doubled,  is  the  intercepting  central  angle  doubled  ? 
Is  the  subtending  chord  also  doubled  ?  .  Why  is  it  that  a  central  angle 
is  measured  by  its  intercepted  arc,  while  a  chord  is  not  measured  by 
its  subtended  arc  ? 


152  PLANE  GEOMETRY— BOOK  II 

PROPOSITION  VII.    THEOREM 

268.  The  shortest  line  and  also  the  longest  line  which  can 
be  drawn  from  a  point  to  a  circle  lie  on  a  line  passing  through 
the  center  of  the  circle  and  the  given  point. 


FIG.  2 

Given  the  circle  0  and  the  point  A,  and  AB  and  AC  the  shorter  and 
longer  segments  respectively  of  a  line  drawn  through  A  and  passing 
through  0,  the  center  of  the  circle. 

To  prove  that  AB  is  the  shortest  line,  and 'AC  the  longest  line,  from  A  to 
the  circle. 

CASE  I.    When  A  is  within  the  circle  (Fig.  1). 

Proof.    1.  Draw  DE,  any  other  line  through  A,  cutting  the  circle 
in  D  and  E.    Let  AE  be  its  longer  and  AD  its  shorter  segment. 
Draw  the  radii  OE  and  OD. 

2.  In  the  A  OAD,  OA  +  AD  >  OD ;        .  Why  ? 
that  is,                                     OA  +  AD  >  OB. 

(Since  OB  =  OD.) 

Or  OA  +  AD  >  OA  +  AB ;  that  is,  AD  >  AB.  Ax.  10 

Hence  AB  is  less  than  any  other  line  from  A  to  the  circle. 

3.  Also  in  the  A  OAE,     OA  +  OE  >  AE ;  Why  ? 
that  is,                        OA  +  OC  >  AE,  or  A  C>  AE.  Why  ? 

Hence  A  C  is  greater  than  any  other  line  from  A  to  the  circle. 

4.  .'.  AB  is  the  shortest  line,  and  AC  the  longest  line,  from  A  to 
the  circle. 

CASE  II.    When  A  is  without  the  circle  (Fig.  2). 

(To  be  completed.) 

269.  The  distance  from  a  point  to  a  circle  is  the  shortest  line  from 
the  point  to  the  circle. 


BOOK  II  153 

TANGENTS 
PROPOSITION  VIII.    THEOREM 

270.  A  straight  line  perpendicular  to  a  radius  at  its  ex- 
tremity is  a  tangent  to  the  circle. 


AT          G 

Given  the  circle  0,  the  radius  OC,  and  the  line  AB  perpendicular 
to  OC  at  its  extremity  C. 

To  prove  that  AB  is  tangent  to.  the  circle  O. 

Proof.    1.  From  the  center  O  draw  any  other  line  to  AB,  as  OT. 

2.  Then  OT>OC.  Why? 

3.  .'.  the  point  T  is  without  the  circle.  Why  ? 

4.  Hence  every  point  in  the  line  AB,  except  C,  lies  without 
the  circle,  and  therefore  AB  is  tangent  to  the  circle  at  C. 

(Definition  of  a  tangent,  §  66.) 

271.  COROLLARY  1.  A  tangent  to  a  circle  is  perpendicular  to 
the  radius  drawn  to  the  point  of  contact. 

For  0(7  is  the  shortest  line  from  0  to  AB,  and  is  therefore  JL  to  AB. 

272.  COROLLARY  2.  A  perpendicular  to  a  tangent  at  the  point 
of  contact  passes  through  the  center  of  the  circle. 

For  the  radius  drawn  to  the  point  of  contact  is  JL  to  the  tangent,  and 
therefore  a  _L  erected  at  the  point  of  contact  coincides  with  this  radius 
and  passes  through  the  center. 

273.  COROLLARY  3.  A  perpendicular  from  the  center  of  a 
circle  to  a  tangent  passes  through  the  point  of  contact. 


154       PLANE  GEOMETRY— BOOK  II 

PROPOSITION  IX.    PROBLEM 

274.    Through  a  given  point,  to  draw  a  tangent  to  a  given 
circle. 


— -B 


FIG.  1  FIG.  2 

Given  the  circle  0  and  the  point  P. 

Required  to  construct  a  tangent  to  the  circle  0  through  the 
point  P. 

CASE  I.     When  the  given  point  P  is  on  the  circle  0  (Fig.  1). 
Construction.    1.      Draw  the  radius  OP. 
2.  Construct  a  line  AB  _L  to  OP  at  P.  §  263 

Then  AB  is  the  tangent  required.  Why  ? 

CASE  II.   When  the  given  point  P  is  ivithout  the  circle  (Fig.  2). 

Construction.    1.  Draw  the  line  OP,  joining  the  given  point 
and  the  center  of  the  O. 

2.  With  OP  as  a  diameter  describe  a  circle  intersecting  the 
given  circle  at  A  and  B. 

3.  Draw  PA  and  PB.    Each  of  these  lines  is  tangent  to  the 
circle  0. 

Proof.    1.  Draw   OA.    Then  the  AOAP,  being  inscribed  in 
a  semicircle,  is  a  right  triangle.  §  262 

2.  .'.PAJLOA.  Why? 

3.  /.  PA  is  a  tangent  to  the  circle  0.  Why  ? 

In  the  same  manner,  by  drawing  the  radius  OB,  it  may  be 
proved  that  the  line  PB  is  a  tangent  to  the  circle  0. 

Discussion.   There  are,  therefore,  two  solutions  of  the  above  problem, 
when  the  given  point  is  an  external  point. 


TANGENTS  155 

PROPOSITION  X.    THEOREM 

275.  The  tangents  drawn  to  a  circle  from  an  external  point 
are  equal,  and  make  equal  angles  with  the  line  joining  that 
point  to  the  center. 


Given  AB  and  AC  tangent  to  the  circle  0;  also  the  line  OA. 
To  prove  that      AB  =  A  C,  and  Z  m  =  Z  n. 
(To  be  completed.) 

276.  The  chord  connecting  the  points  of  contact  of  the  two 
tangents  to  a  circle  from  an  external  point  is  called  the  chord 
of  contact  of  the  tangents. 

277.  A  circle  is  said  to  be  inscribed  in  a  given  polygon  when 
the  sides  of  the  polygon  are  tangent  to  the  circle.   The  polygon 
is  said  to  be  circumscribed  about  the  circle. 

EXERCISES 

1.  In  the  above  figure 

What  would  be  the  relation  of  OA  to  the  chord  of  contact  BCt 

Could  the  chord  B  C  under  any  circumstances  be  the  perpendicular 
bisector  of  OA  ? 

What  relation  exists  between  ZBOC  and  ZBAC1? 

If  &BAC  is  equilateral,  how  many  degrees  in  the  ZBOCt 

If  /.BAG  is  120°,  prove  that  AB  +  A  C  =  OA. 

Prove  that  the  quadrilateral  ABOC  can  be  inscribed  in  a  circle 
having  OA  for  its  diameter. 

If  OA  =  2  OB,  how  many  degrees  in  the  Z7L4C? 

2.  Construct  a  tangent  to  a  given  circle  parallel  to  a  given  line ; 
perpendicular  to  a  given  line.    How  many  solutions  are  there? 


156       PLANE  GEOMETRY— BOOK  II 

PROPOSITION  XI.    PROBLEM 
278.    To  inscribe  a  circle  in  a  given  triangle. 


A 


o  D 

Given  the  triangle  ABC. 
Required  to  inscribe  a  circle  in  A  ABC. 

Construction.    1.  Bisect  the  angles  B  and  C. 

2.  From  0,  the  intersection  of  the  bisectors,  draw  OD  _L  to  BC. 

3.  About  0  as  a  center,  with  a  radius  equal  to  OD,  draw  a  O. 
This  is  the  O  required. 

Proof.    1.   0  is  equidistant  from  AB,  AC,  and  BC.  §  245 

2.  .'.  ABjAC,  and  BC  are  each  _L  to  a  radius  of  the  above  O 
at  its  extremity.  Why  ? 

3.  .'.  ABj  ACj  and  BC  are  tangents  to  the  O ;  that  is,  the  O  is 
inscribed  in  the  A  ABC.  §  277 

EXERCISES 

1.  If  two   circles  are  concentric,  all  chords   of  the  outer  circle 
which  are  tangents  to  the  inner  are  equal  to  each  other  and  are 
bisected  at  their  points  of  contact. 

2.  A  circle  can  be  inscribed  in  a  rhombus. 

3.  The  sum  of  two  opposite  sides  of  a  quadrilateral  circumscribed 
about  a  circle  is  equal  to  the  sum  of  the  other  two  sides. 

4.  Conversely,  if  -the. sum  of  two  opposite  sides  of  a  quadrilateral 
is  equal  to  the  sum  of  the  other  two  sides,  then  a  circle  may  be  in- 
scribed in  the  quadrilateral.    (Prove  by  constructing  a  circle  tangent 
to  three  sides  of  the  quadrilateral,  and  showing  that  if  the  fourth 
side  were  to  cut  the  circle  or  to  lie  wholly  without  the  circle,  it 
would  lead  to  an  absurdity.) 


TANGENTS 


157 


5.  Given  a  circumscribed  hexagon ;  prove  that  the  sum  of  one 
set  of  alternate  sides  (first,  third,  fifth)  equals  the  sum  of  the  other 
set  (second,  fourth,  sixth). 

6.  Is  the  statement  of  Ex.  5  true  for  other  circumscribed  polygons? 

7.  The  radius  of  a  circle  inscribed  in  an  equilateral  triangle  is 
equal  to  one  third  of  the  altitude. 

8.  The  hypotenuse  of  a  right  triangle  is  equal  to  the  sum  of  the 
legs  diminished  by  twice  the  radius  of  the  inscribed  circle. 

9.  A  parallelogram  circumscribed  about  a  circle  is  either  a  rhom- 
bus or  a  square. 

10.  If  from  the  extremities  of  a  diameter  of  a  circle  perpen- 
diculars are  dropped  upon  a  tangent,  the  sum  of  the  perpendiculars 
is  equal  to  the  diameter. 

11.  When  a  ray  of  light  strikes  a  spherical  mirror  (represented  in 
cross  section  by  a  circle),  the  angle  of  incidence  (see  Ex.  32,  p.  107) 
is  found  by  drawing  a  tangent  to  the  circle  at  the  point  of  incidence, 
and  erecting  a  perpendicular  to  the  tangent  at  that 

point.    In  this  case  the  perpendicular  (called  the 
normal)  is  a  radius.    (Why?) 

The  line  of  propagation  of  a  sound  wave  also 
follows  the  law  of  reflection  of  a  ray  of  light, 
namely,  that  the  angle  of  incidence  is  equal  to 
the  angle  of  reflection. 

The  circular  gallery  in  the  dome  of  St.  Paul's  in 
London  is  known  as  a  whispering  gallery,  for  the 
reason  that  a  faint  sound  produced  at  a  point  near  the  wall  can  be 
heard  around  the  gallery  near  the  wall,  but  not  elsewhere.  The 
sound  is  reflected  along  the  circular  wall  in  a  series  of  equal  chords. 
Explain  why  these  chords  are  equal. 

12.  Two  straight  roads  of  different  width  meet 
at  right  angles.    It  is  desired  to  join  them  by  a 
road  the  sides  of  which  are  arcs  of  circles  tangent 
to  the  sides  of  the  straight  roads.  What  construc- 
tion lines  are  necessary?   Draw  such  a  figure. 

13.  A  circle  tangent  to  one  side  of  a  triangle 
and  to  the  other  two  sides  produced  is  called  an 

escribed  circle.  How  many  escribed  circles  has  a  triangle  ?  Draw  them. 


158 


PLANE  GEOMETEY— BOOK  II 


PROPOSITION  XII.    THEOREM 
279.  Parallel  lines  intercept  equal  arcs  on  a  circle. 


FIG.  1 


FIG.  2 


CASE  I.     When  the  parallels  are  a  tangent  and  a  secant. 

.Given  the  circle  0,  with  AB  (Fig.  1),  a  tangent  at  F,  parallel  to 
CD,  a  secant. 

To  prove  that  arc  CF  =  arc  DF. 

Proof.    1.  Draw  the  radius  OF. 

2.  Then  OF±AB.  Why? 

3.  And  hence  OF  is  also  _L  to  CD.  Why  ? 

4.  .'.  arc  CF  =  arc  DF.  Why  ? 

(Case  II,  when  both  parallels  are  secants,  and  Case  III,  when  both  are 
tangents,  are  left  as  exercises.) 

EXERCISES 

1.  A  trapezoid  inscribed  in  a  circle  is  isosceles. 

2.  A  circle  has  two  parallel  chords,  which  lie  on  opposite  sides  of 
the  center.    One  is  equal  to  the  side  of  a  regular  hexagon,  the  other 
to  the  side  of  a  regular  decagon  inscribed  in  the  given  circle.    Find 
the  number  of  degrees  in  the  arcs  intercepted  between  the  chords. 

3.  Solve  Ex.  2,  if  the  two  chords  are  respectively  sides  of  a  penta- 
gon and  an  octagon ;  a  pentagon  and  a  dodecagon ;  a  hexagon  and 
an  octagon ;  a  square  and  a  decagon ;  a  triangle  and  a  hexagon. 

4.  Solve  Exs.  2  and  3,  if  the  parallel  chords  are  on  the  same  side 
of  the  center. 


BOOK  II  159 

TWO  CIRCLES 
PROPOSITION  XIII.    THEOREM 

280.  If  two  circles  intersect  each  other,  the  line  of  centers  is 
perpendicular  to  their  common  chord  at  its  middle  point. 


Given  two  circles  whose  centers  are  C  and  Cf,  intersecting  at  the 
points  A  and  £,  the  common  chord  AB,  and  the  line  of  centers  CCf. 

To  prove  that  CC1  is  _L  to  AB  at  its  middle  point. 
Proof.    1.          Draw  CA,  CB,  C'A,  and  C'B. 

2.  Then  CA  =  CB,  and  C'A  =  C'B.  Why  ? 

3.  .'.  CC'  is  the  perpendicular  bisector  of  AB.      Why? 

281.  Two  circles  are  tangent  to  each  other  if  both  are  tangent 
to  a  straight  line  at  the  same  point.  The  circles  are  said  to  be 
tangent  internally  or  externally,  according  as  one  circle  lies 
wholly  within  or  without  the  other.  The  common  point  is  called 
the  point  of  contact. 

EXERCISES 

1.  Describe  the  relative  position  of  two  circles,  if  the  line-segment 
joining  their  centers  is  equal  to  the  sum  of  the  radii ;  is  equal  to  the 
difference  of  the  radii. 

2.  Given  the  line  Z  and  the  point  P  on  that  line.    Construct  two  cir- 
cles tangent  to  Z  at  P,  each  with  a  radius  equal  to  another  line  m. 

3.  (riven  two  intersecting  lines  and  a  point  P  on  one  of  those  lines. 
Construct  two  circles  tangent  to  both  lines  and  passing  through  P. 


160       PLANE  GEOMETRY— BOOK  II 

PROPOSITION  XIV.    THEOREM 

282.  If  two  circles  are  tangent  to  each  other,  the  line  of 
centers  passes  through  the  point  of  contact. 


Given  the  circles  C  and  C'  tangent  to  the  straight  line  AB  at  0, 
and  the  line  of  centers  CCr. 

To  prove  that  the  line  CC'  passes  through  0. 

Proof.  1.  A  line  _L  to  AB  at  the  point  0  will  pass  through 
the  centers  C  and  C'.  Why  ? 

2.  Hence  the  line  CC'  must  coincide  with  that  _L       Why  ? 

3.  .'.  0  is  in  the  straight  line  CC'. 

EXERCISES 

1.  PROBLEM.    To  construct  three  circles  about  the  vertices  of  a 
given  triangle  as  centers,  each  circle  being  tangent  externally  to  the 
other  two. 

Suggestion.    Inscribe  a  circle  in  the  given  triangle. 

2.  Find  the  radii  of  three  such  circles,  if  the  sides  of  the  triangle 
are  4,  5,  and  6  (8,  8,  and  14). 

3.  In  accurate  tool  work,  where  holes  are  to  be  bored  near  each 
other  in  a  metal  plate,  the  centers  are  first  marked  carefully  to 
thousandths  of  an  inch.    This  is  often  done  by  first  turning  disks 
on  a  lathe,  the  diameters  of  the  disks  being  such  that  when  they  are 
placed  tangent  to  each  other  their  centers  will  mark  the  positions 
of  the  centers  of  the  holes  to  be  bored. 

Three  holes  are  to  be  bored.  The  distances  between  their  centers  are 
0.650  in.,  0.790  in.,  and  0.865  in.  respectively.  Find  the  diameters  of 
the  required  disks.  (From  "  School  Science  and  Mathematics.") 


TWO  CIRCLES 

PROPOSITION  XV.    PROBLEM 
283.    To  draw  a  common  tangent  to  two  given  circles. 


161 


CASE  I.  When  the  tangent  is  to  cross  the  line-segment  joining  the  centers 
(common  internal  tangent). 

Given  the  circles  0  and  0'. 

Required  to  construct  a  common  internal  tangent  to  the  circles  0  and  0'. 

Analysis.  It  is  apparent  that  such  a  tangent,  when  drawn,  would 
be  _L  to  the  radii  OM  and  O'-ZV  drawn  to  M  and  N,  the  points  of  tan- 
gency.  Now  if  through  0'  a  line  be  drawn  II  to  the  common  tangent, 
cutting  OM  produced  in  P,  MNO'P  would  be  a  rectangle,  MP  would 
be  equal  to  NO',  and  OP  would  be  equal  to  OM  +  O'N,  or  the  sum  of 
the  radii ;  also,  O'P  would  be  tangent  to  a  O  about  0  with  a  radius 
equal  to  OP,  that  is,  equal  to  OM  +  O'N. 

(Construction  and  proof  to  be  completed.) 


CASE  II.  When  the  tangent  is  not  to  cross  the  line-segment  joining  the 
centers  (common  external  tangent}. 

Analysis.  The  conditions  of  the  problem  in  this  case  are  identical 
with  those  of  Case  I,  except  that  the  radius  OP  will  be  equal  to  the 
difference  of  the  radii  OM  and  &N 

(Construction  and  proof  to  be  completed.) 


162 


PLANE  GEOMETKY— BOOK  II 


EXERCISES 

1.  If  two  common  external  tangents  or  two  common  internal  tan- 
gents are  drawn  to  two  circles,  the  segments  intercepted  between  the 
points  of  contact  are  equal. 

2.  How  should  two  wheels  be  connected  (a)  by  belting,  and  (b)  by 
gearing,  so  as  to  revolve  in  the  same  direction  ?  in  opposite  directions  ? 

3.  In  each  of  the  different  positions  of  two  circles  mentioned  in 
Ex.  1,  p.  39,  how  many  common  internal  and  how  many  common 
external  tangents  may  be  drawn  ? 

4.  Construct  the  common  tangents  to  two  equal  circles  which  do 
not  intersect. 

5.  About  the  vertices  of  an  equilateral  triangle  as  centers,  with 
radii  equal  to  one  half  the  side  of  the  triangle, 

draw  circles.  These  circles  are  tangent  to  each 
other.  (Why  ?)  Show  how  to  construct  a  circle 
which  shall  inclose  these  circles  and  be  tangent 
to  all  three  of  them.  (This  construction  is  the 
basis  of  a  design  which  appears  frequently  in 
decoration.) 

6.  Construct  a  circle  which  shall  be  tangent 
to  a  given  arc,  and  to  the  radii  drawn  to  the 
ends  of  the  arc. 

(If  a  tangent  is  drawn  at  the  mid-point  of 
the  arc,  show  that  the  required  circle  will  be 
inscribed  in  the  triangle  formed  by  this  tangent 
and  the  radii  produced.) 

7.  Arcs  of  circles,  either  alone  or  combined  with  their  chords,  are 
used  to  form  arches,  which  appear  frequently  in 

decorative  design.    A  common  form  is  the  semicir- 
cular arch,  which  consists  of  a  semicircle  and  its 
diameter.    Inscribe   a  circle  in   such   an  arch   as 
shown  in  the  figure.    Would  the  same  method  of  construction  hold 
good  if  the  arc  were  less  than  a  semicircle  ? 

8.  Apply  Ex.  6,  (a)  to  inscribe  in  a  given  semicircular  arch  two 
equal  tangent  circles ;  (b)  to  draw  in  a  semicircular  arch  three  equal 
circles  tangent  to  the  semicircle,  of  which  two  shall  be  tangent  to  the 
diameter,  and  the  third  tangent  to  the  other  two  circles. 


TWO  CIRCLES 


163 


9.  The  spiral  shown  in  Fig.  1  is  composed  of  semicircles,  having 
as  centers  the  points  A  and  B  alternately.  Draw  such  a  spiral,  and 
prove  that  two  consecutive  semicircles  are  tangent  to  each  other. 


FIG.  1 

10.  A  four-centered  spiral  may  be  constructed  by  extending  in 
succession  the   sides  of  a  square   and   using   each  vertex   as   the 
center  of  a  quadrant,  as  shown  in  the  figure.    Draw  such  a  spiral 
(Fig.  2). 

11.  Construct  in  similar  fashion  a  three-centered  spiral  of  three 
arcs ;  a  six-centered  spiral  of  six  arcs. 

12.  Three  circles  are  drawn  in  an  equilateral 
triangle,  such  that  each  is  tangent  to  the  other 
two  circles  and  to  two  sides  of  the  triangle.   Show 
that  each  circle  is  tangent  to  two  altitudes  of  the 
triangle,  and  give    a    method    for  making  the 
construction. 

13.  Within  a  given  square  construct  four  equal  circles  such  that 
each  is  tangent  to  two  others  and  to  two  sides  of  the  square. 

(Divide  the  square  into  four  smaller  squares  and  inscribe  a  circle 
in  each.    Show  that  these  circles  answer  the  required  conditions.) 

14.  Within   a    given    square    construct    four 
equal  circles  such  that  each  is  tangent  to  two 
others  and  to  one  side  of  the  square. 

(Draw  the  diagonals  of  the  square.) 

15.  Construct  the  design  shown  in  the  figure. 
(Note  that  the  radius  of  each  of  the  small 

circles  is  one  third  the  radius  of  the  large  circle. 

Three  of  the  small  circles  lie  along  one  diameter.    How  are  the 

centers  of  the  other  four  small  circles  found?) 


164 


PLANE  GEOMETRY— BOOK  II 


ANGLE  MEASUREMENTS 

284.  An  angle  is  inscribed  in  a  circle  when  its  vertex  is  on  the 
circle  and  its  sides  are  chords. 

An  angle  is  inscribed  in  an  arc  when  its  vertex  lies  on  the  arc 
and  its  sides  pass  through  the  ends  of  the  arc. 

PROPOSITION  XVI.    THEOREM 

285.  An  inscribed  angle  is  measured  by  one  half  the  arc 
intercepted  between  its  sides. 


FIG.  3 

Given  the  circle  0,  with  the  angle  BAG  inscribed  in  it. 
To  prove  that     Z.BAC  is  measured  by  %  arc  EC. 
CASE  I.    When  the  center  0  is  in  one  of  the  sides  of  the  angle 
(Fig-  1). 

Proof.    1.  Draw  OC. 

2.  Then  ZA  =  Z.C.  Why? 

3.  But  Z.BOC  =  £A+Z.C.  Why? 


4.  But  Z.EOC  is  measured  by  arc  EC.  §94 

(A  central  angle  is  measured  by  its  intercepted  arc.) 

5.  .'.  /-A  is  measured  by  \  arc  EC.  Ax.  5 

(Cases  II  and  III,  when  the  center  falls  within  or  without  the  angle, 
to  be  completed.) 

REMARK.  This  theorem  is  otherwise  stated  :  "  The  number  of  degrees 
in  an  inscribed  angle  is  equal  to  one  half  the  number  of  arc-degrees  in 
the  intercepted  arc  "  (see  §  92). 


ANGLE  MEASUREMENTS 


165 


286.  COROLLARY  1.  An  angle  inscribed  in  a  semicircle  is  a 
right  angle. 

For  it  is  measured  by  half  a  semicircle  (Fig.  4).    See  also  §  262. 

287.  COROLLARY  2.  An  angle  inscribed  in  an  arc  greater  than 
a  semicircle  is  an  acute  angle. 

For  it  is  measured  by  one  half  an  arc  less  than  a  semicircle ;  as  Z.A 
(Fig.  5). 


FIG.  4 


FIG.  5 


288.  COROLLARY  3.  An  angle  inscribed  in  an  arc  less  than  a 
semicircle  is  an  obtuse  angle. 

289.  COROLLARY  4.  Angles  inscribed  in  the  same  arc,  or  in 
equal  arcs  of  the  same  circle,  are  equal  (Fig.  6). 


EXERCISES 

1.  Test  by  Proposition  XVI  the  following  theorems  : 

(a)  The    sum    of    the    angles    of    a    triangle    equals    two    right 
angles. 

(b)  If  two  sides  of  a  triangle  are  equal,  the  angles  opposite  are 
equal. 

(c)  If  two  angles  of  a  triangle  are  equal,  the  sides  opposite  are 
equal. 

These  tests  cannot  be  regarded  as  proofs.    Why  ? 

2.  If  two  triangles  have  their  angles  respectively  equal  and  are 
inscribed  in  the  same  circle,  they  are  congruent. 

3.  The    bisectors   of    all    angles  inscribed   in  the   same   arc   are 
concurrent. 


166       PLANE  GEOMETRY— BOOK  II 

PROPOSITION  XVII.    THEOREM 

290.  An  angle  included  by  a  tangent  and  a  chord  drawn 
from  the  point  of  contact  is  measured  by  one  half  the  inter- 
cepted arc. 


c 


B 


Given  the  angle  m  included  by  AB,  a  tangent  to  the  circle  0  at 
C,  and  the  chord  CD. 

To  prove  that    /-  m  is  measured  by  \  arc  CD. 
Proof.    1.  Draw  DE  II  to  AB  through  D. 

arc  CD  =  arc  CE. 

Z.m  =  /-  n. 

/.  n  is  measured  by  ^  arc  CE. 
.  Z.  m  is  measured  by  \  arc  CD. 
is  measured  by  1  major  arc  CED. 


2.  Then 

3.  Also 

4.  But 

5.  Also 


Why? 
Why  ? 
Why? 
Ax.  1. 


(Being  the  supp.  of  Z  m,  while  arc  CED  is  the  conjugate  of  arc  CD.) 


EXERCISES 

1.  The  opposite  angles  of  an  inscribed  quadrilateral  are  supple- 
mentary. 

(By  which  arcs  are  A  A  and  C  measured? 
What  is  the  sum  of  these  arcs?) 

2.  If  the  opposite  angles  of  a  quadrilateral 
are  supplementary,  the  quadrilateral   may  be 
inscribed  in  a  circle  (converse  of  Ex.  1). 

Suggestion.   Pass  a  circle  through  three  of  the 
vertices  (§  258),  and  show  by  the  Indirect  Method  that  the  fourth 
vertex  must  lie  on  the  circle. 


ANGLE  MEASUREMENTS 
PROPOSITION  XVIII.    THEOREM 


167 


291 .  An  angle  formed  by  two  chords  intersecting  within  the 
circle  is  measured  by  half  the  sum  of  the  intercepted  arcs. 


Given  the  circle  0,  in  which  two  chords,  AB  and  CD,  intersect  in  E. 
To  prove  that  /.AED  is  measured  by  ^  (arc  AD  +  arc  CB). 
Proof.    1.  Draw  DB. 

2.  Then  Z.AED  =  ^B  +  Z7X  Why? 

(To  be  completed.) 

PROPOSITION  XIX.    THEOREM 

292.  An  angle  formed  by  two  secants,  or  two  tangents,  or  a 
tangent  and  a  secant,  intersecting  without  a  circle,  is  measured 
by  half  the  difference  of  the  intercepted  arcs. 


Given  the  circle  0  with  two  secants,  or  two  tangents,  or  a  tangent 
and  a  secant  intersecting  at  A  without  the  circle. 

To  prove  that  Z.A  is  measured  by  half  the  difference  of  the 
intercepted  arcs. 

(Cases  I,  II,  and  III  to  be  completed.) 


168       PLANE  GEOMETRY— BOOK  II 

PROPOSITION  XX.    PROBLEM 

293.    Upon  a  given  straight  line  as  a  chord  to  construct  an 
arc  of  a  circle  in  which  a  given  angle  may  be  inscribed. 

E  „ 


Given  the  straight  line  AB  and  the  angle  R. 

Required  to  construct,  on  the  line  AB  as  a  chord,  an  arc  of  a 
circle  such  that  the  angle  R  may  be  inscribed  in  the  arc. 

Construction.    1.     Construct  the  Z.ABC  equal  to  Z.R. 

2.  Draw  DP  the  _L  bisector  of  AB. 

3.  At  B  construct  BM 1.  to  EC. 

4.  About  0,  the  point  of  intersection  of  DP  and  BM,  as  a 
center,  with  a  radius  equal  to  OB,  draw  a  circle.    This  circle 
will  pass  through  A,  and  the  arc  A EB  is  the  arc  required. 

Proof.    1.    The  point  0  is  equidistant  from  A  and  B.    Why  ? 
. ' .  the  circle  passes  through  A .  Why  ? 

2.  But  BC±.BO.  Cons. 

.'.  EC  is  tangent  to  the  O.  Why  ? 

3.  Then       /.ABC  is  measured  by  £  arc  AFB.  Why  ? 

4.  But  any  angle,  as  Z.AEB,  inscribed  in  the  arc  AEB,  is 
also  measured  by  £  arc  AFB.  Why? 

5.  .'.  any  angle  inscribed  in  the  arc  AEB  equals  /.R. 


ANGLE  MEASUREMENTS  169 

NUMERICAL  EXERCISES 

1.  An  inscribed  angle  intercepts  an  arc  of  40°  (50°,  60°,  75°). 
How  many  degrees  in  the  angle  ? 

2.  An  angle  of  20°  (30°,  45°,  50°,  67°  30')  is  inscribed  in  a  circle. 
How  many  degrees  in  the  intercepted  arc  ? 

3.  The  sides  of  an  inscribed  triangle  subtend  arcs  of  100°,  120°, 
and  140°.    How  many  degrees  in  each  angle  of  the  triangle  ? 

4.  The  arcs  subtended  by  the  sides  of  an  inscribed  triangle  are 
in  the  ratio  1:2:3.    What  kind  of  triangle  is  it? 

5.  The  sides  of  an  inscribed  quadrilateral  subtend  arcs  in  the 
ratio  1:2:3:4  (3:5:7:  9).    How  many  degrees  in  each  angle  of 
the  quadrilateral  ? 

6.  In  the  preceding  exercise  how  many  degrees  in  the  angles 
between  the  diagonals  of  the  quadrilateral  ? 

7.  A  triangle  is  inscribed  in  a  circle,  and  another  triangle  is  cir- 
cumscribed by  drawing  tangents  at  the  vertices  of  the  inscribed  tri- 
angle.   The  angles  of  the  inscribed  triangle  are  40°,  60°,  and  80°. 
Find  all  the  other  angles  of  the  figure. 

8.  The  vertex  angle  of  an  inscribed  isosceles  triangle  is  100°. 
How  many  degrees  in  the  arcs  subtended  by  each  of  the  sides  ? 

9.  The  bases  of  an  inscribed  isosceles  trapezoid  subtend  arcs  of 
100°  and  120°.    How  many  degrees  in  each  angle  of  the  trapezoid 
(a)  if  the  bases  are  on  the  same  side  of  the  center?  (b)  if  they  are 
on  opposite  sides  of  the  center  ? 

10.  At  the  vertices  of  an  inscribed  quadrilateral  tangents  are 
drawn  to  the  circle,  forming  a  circumscribed  quadrilateral.  The  arcs 
subtended  by  the  sides  of  the  inscribed  quadrilateral  are  in  the  ratio 
3:4:5:8.  Find 

(a)  the  angles  of  each  quadrilateral ; 

(b)  the  angles  between  the  diagonals  of  the  inscribed  quadri- 
lateral ; 

(c)  the  angles  between  the  opposite  sides  of  the  inscribed  quad- 
rilateral produced  to  intersect ; 

(d)  the  angles  between  the  sides  of  the  inscribed  and  those  of 
the  circumscribed  quadrilateral. 


170 


PLANE  GEOMETRY— BOOK  II 


11.  A  regular  hexagon  is  inscribed  in  a  circle  and  all  its  diagonals 
are  drawn.    How  many  degrees  in  each  angle  of  the  figure  ? 

12.  The  angle  formed  by  two  tangents  drawn  to  the  same  circle 
is  100°.    How  many  degrees  in  the  two  arcs  subtended  by  the  chord 
of  contact? 

13.  If  the  angle  between  two  tangents  to  the  same  circle  is  60°, 
what  kind  of  triangle  is  formed  when  the  chor.d  of  contact  is  drawn  ? 

14.  Given  two  tangents  to  a  circle ;  the  major  arc  contains  200°. 
How  many  degrees  in  the  angle  formed  by  the  tangents  ? 

15.  The   sides  of   an  inscribed  pentagon   subtend   arcs  each   of 
which  is  10°  greater  than  the  preceding  one.    Draw  the  diagonals 
of  the  pentagon  and  determine  each  angle  of  the  figure. 

16.  In  the  preceding  exercise  draw  tangents  at  the  vertices  of  the 
inscribed  pentagon  and  determine  each  of  the  interior  angles  of  the 
resulting  circumscribed  pentagon. 

17.  The  diagram  shows  how  the  latitude  of 
a  place  may  be  determined  by  observation  of 
the  pole  star.    Let  EE'  represent  the  equator, 
A  A'  the  axis  of  the  earth,  P  the  place  whose 
latitude  is  to  be  found,  PF  a  plane  tangent  to 
the  earth  at  P  (the  horizon),  and  PS  the  line 
of  observation  of  the  pole  star.    Then  Z.  of  rep- 
resents the  latitude,  and  Z.  a    is  called  the  alti- 
tude of  the  pole  star.  For  practical  purposes  we 
may  assume  that  A  A'  II  PS. 

Prove  that  Z  a'  =  Z  a. 

18.  An  inscribed  angle  is  formed  by  the  side  of  a  regular  inscribed 
hexagon  and  the  side  of  a  regular  inscribed  decagon.    How  many 
degrees  in  the  angle  ?    (Give  two  solutions.) 

19.  Solve  Ex.  18  if  the  inscribed  angle  is  formed  by  sides  of  the 
following  regular  polygons  : 

(a)  triangle  and  square  ; 

(b)  pentagon  and  octagon  ; 

(c)  hexagon  and  octagon ; 

(d)  pentagon  and  dodecagon. 


BOOK  II  171 

LOCI 

294.  The  definition  of  a  circle  may  be  used  to  introduce  a 
very  important  geometric  idea.  All  points  in  a  plane  which  are 
two  inches  from  a  given  point  0  lie  in  a  circle 
whose  center  is  the  given  point  0  and  whose 
radius  is  two  inches  long. 


Conversely,  every  point  in  this  circle  is  two 
inches  from  0. 

These   two    statements    are    sometimes    re- 
placed by  the  one  statement  that  the  circle  about  0  is  the  locus 
(that  is,  place]  of  the  points  two  inches  from  0.    In  general, 

The  locus  of  a  point  satisfying  a  given  condition  is  the  figure 
containing  all  the  points  that  fulfill  the  given  condition  (or 
answer  the  given  description),  and  no  other  points. 

Hence  the  following 

295.  Rule  for  solving  Locus  Problems  : 

1.  Locate  a  number  of  points  which  satisfy  the  given  con- 
dition, and  thus  obtain  a  notion  of  what  the  locus  is. 

2.  Prove  that  every  point  satisfying  the  given  condition  lies 
in  the  assumed  locus. 

3.  Prove  'that  every  point  of  the  assumed  locus  satisfies  the 
given  condition. 

In  the  following  simple  exercises,  however,  the  answers  may 
be  stated  without  proof. 

EXERCISES 

1.  Where  are   all  the   houses  that  are  1  mi.  from  your   school 
building  ? 

2.  Where  are  all  the  villages  that  are  10  mi.  from  your  own  town  ? 

3.  Sound  travels  about  1100  ft.  per  second.    If  a  cannon  is  fired 
from  a  certain  point,  what  is  the  locus  of  all  persons  who  hear  the 
report  after  3  sec.  ? 

4.  A  ladder  leans  against  a  wall.    A  man  stands  on  the  middle 
rung.    If  the  ladder  slips  down,  what  is  the  locus  of  the  man's  feet? 


172 


PLANE  GEOMETRY— BOOK  II 


5.  What  locus  problems  are  suggested  to  you  by  the  opening  of  a 
book  or  a  door  ?  by  a  seesaw  ?  a  pendulum  ?  the  governor  of  an 
engine  ?  a  clock  ? 

Theorem  I.  The  locus  of  a  point  A  at  a  given  distance !  d 
from  a  given  point  P  is  a  circle  having  P  as  its  center  and  d 
as  its  radius. 

EXERCISES 

1.  What  is  the  locus  of  a  house  100  ft.  from  a  straight  road? 

2.  On  a  city  map  what  is  the  locus  of  a  house  200  ft.  from  the 
"  mile  circle  "  ? 

3.  Where  are  all  the  points  that  are  2  in.  from  the  surface  of  the 
table  ? 

4.  What  is  the  locus  of  the  foot  of  a  tree  which  is  10  ft. 'from  the 
wall  of  a  round  tower? 

5.  What   is  the  locus  of  a  point  1  in.  from  the  surface  of   a 
spherical  shell? 

Theorem  II.    The  locus  of  a  point  A  at  a  given  distance  d  from 
a  given  line  I  consists  of  two  lines 
parallel  to  I,  one  on  each  side  of          ^ 
I,  and  d  units  from  it  (§  237). 

Theorem  III.  The  locus  of  a  \d_d_ 
point  A  at  a  given  distance  d  from 
a  given  circle  whose  radius  is  R 
consists  of  two  circles  concentric 
with  the  given  circle  and  with 
radii  R  +  d  and  R  —  d  respectively  (§  269). 


EXERCISES 

1.  A  motor  boat  sails  up  a  straight  canal  midway  between  the 
banks.    What  is  the  locus  of  the  boat? 

2.  The  hands  of  a  clock  describe  concentric  circles.    What  is  the 
locus  of  a  point  equidistant  from  these  circles  ? 

3.  What  is  the  locus  of  a  point  equidistant  from  two  opposite 
walls  of  a  rectangular  room  ? 


LOCI 


173 


Theorem  IV.  The  locus  of  a  point  equidistant  from  two  parallel 
lines  is  a  parallel  line  midway  be-        [ 
tween  the  two  parallels  (§  237). 

Theorem  V.  The  locus  of  a  point 
equidistant  from  two  concentric  cir- 
cles is  a  concentric  circle  midway 
between  those  two  circles  (§  269). 

"What  is  the  radius  of  the  locus  if  the 
radii  of  the  given  circles  are  rx  and  r2  ? 

EXERCISES 

1.  The  poles  of  a  telephone  line  are  to  be  each  equidistant  from 
two  houses.    What  is  the  locus  of  these  poles  ? 

2.  AVhat  is  the  answer  in  Ex.  1,  if  the  poles  of  the  telephone  line 
are  to  be  equidistant  from  two  intersecting  straight  roads  ? 

Theorem  VI.  The  locus  of  a  point  equidistant  from  two 
given  points  is  the  perpendicular  bisector  of  the  line  joining 
the  two  points  (§  239). 

Theorem  VII.  The  locus  of  a  point  equidistant  from  two  in- 
tersecting straight  lines  consists  of  the  pair  of  straight  lines 
that  bisect  the  angles  formed  by  the  two  given  lines  (§  240). 

EXERCISES 

1.  In  front  of  a  rectangular  house,  trees  are  planted  in  such  a  man- 
ner that  the  lines  joining  the  foot  of  each  tree  to  the  two  nearest 
corners  of  the  house  inclose  a  right  angle.  Where  are  the  trees  situated? 

2.  How  would  the  trees  be  located,  if  the  lines  joining  the  foot  of 
each  tree  to  the  two  nearest  corners  of  the  house  inclosed  an  angle  of  40°  ? 

3.  Given  two  fixed  points  A  and  B.    Through  A  a  line  is  drawn 
perpendicular  to  a  line  passing  through  B.    Call  their  point  of  inter- 
section C.   What  is  the  locus  of  C? 

4.  State  the  locus  theorems  underlying  Exs.  1~3  (§  293). 

296.  Intersection  of  loci.  A  point  sometimes  fulfills  more  than 
one  condition.  The  solution  then  consists  in  finding  the  points 
common  to  two  or  more  loci. 


174       PLANE  GEOMETRY— BOOK  II 

EXAMPLE 

Find  a  point  .Y,  a  inches  from  a  given  point  A  and  b  inches 
from  a  given  line  L 

The  solution  is  readily  obtained  by  applying  the  Theorems 
I  and  II  on  page  172.     The  points   of 
intersection  of  the  line  l^  and  the  circle         ,* — * 
in  the  figure  evidently  satisfy  both  con-     /      aj 
ditions.  !          ' 

\  J\^ 

Discussion.   What  is  the  result  (1)  if  ^      \ 
touches  the  circle  ?    (2)  if  the  circle  cuts 
\  and  /,  ?    (3)  if  A  is  on  I  ? 

In  what  way  do  the  relative  values  of  a  and  b  affect  the  result  ? 

EXERCISES 

1.  Construct  a  point  2  in.  from  a  given  point  and  3  in.  from  a 
given  line.    When  is  the  solution  impossible  ? 

2.  Find  a  point  P  which  lies  in  a  given  line  and  is  equidistant 
from  two  given  points.    Discuss  the  problem. 

3.  Construct  a  point  equidistant  from  two  intersecting  lines  and 
also  at  a  given  distance  from  one  of  the  lines. 

4.  Construct  a  point  equidistant  from  the  sides  of  an  angle  and 
at  a  given  distance  from  the  vertex. 

5.  A  tree  is  to  be  planted  10  ft.  from  the  front  wall  and  15  ft.  from 
a  corner  of  a  rectangular  house.    How  many  solutions  are  possible  ? 

6.  Find  in  a  given  circle  a  point  which  is  equidistant  from  two 
given  points. 

7.  Construct  a  point  equidistant  from  two  given  concentric  circles 
and  also  equidistant  from  two  given  lines. 

8.  Find  the  locus  of  the  center  of  a  circle  which  (a)  passes  through 
two  given  points  ;  (b)  touches  each  of  two  given  parallels ;  (c)  touches 
a  given  line  AB  at  a  given  point  P. 

9.  Construct  a  circle  which  has  its  center  in  a  given  line  and 
passes  through  two  given  points. 

10.  Construct  a  circle  which  passes  through  two  given  points  and 
has  its  center  equidistant  from  two  given  intersecting  lines. 


LOCI  175 

11.  Construct  a  point  which  is  at  a  given  distance  from  a  given 
circle  and  at  a  given  distance  from  a  secant  of  that  circle. 

12.  Construct  a  point  which  is  equidistant  from  two  given  parallel 
lines  and  from  a  transversal  to  those  lines. 

13.  Construct  a  circle  which  is  tangent  to  a  given  line  at  a  given  point 
and  has  its  center  on  a  given  line. 

14.  In  the  diagram  let  ABCD  rep- 
resent a  baseball  diamond  (square) ; 
It,  a  street ;  H,  a  house ;  E,  a  corner 
of  the  house  ;  S  and  T,  two  trees ; 
F,  a  fence ;  P,  a  circular  pond.  How 
would  you  locate  a  person  who  is 

(a)  equidistant  from  S  and  T  and  5  yd.  from  CD  ?  (b)  10  yd.  from 
R  and  3  yd.  from  E  ?  (c)  3  ft.  from  P  and  30  yd.  from  0  ?  (d)  equi- 
distant from  R  and  Fl  (e)  equidistant  from  AB  and  C7),  and  from 
S  and  T1  (f)  20  yd.  from  AB  and  5  ft.  from  F? 

MISCELLANEOUS  EXERCISES  ON  LOCI 

1.  What  is  the  locus  of  the  vertex  of  a  triangle  that  has  a  given 
base  and  a  given  altitude  ? 

2.  What  is  the  locus  of  the  vertex  of  the  right  angle  of  a  right 
triangle  which  has  a  given  hypotenuse  ? 

3.  Find  the  locus  of  the  vertex  of  a  triangle  which  has  a  given 
base  and  a  given  vertex  angle. 

4.  What  is  the  locus  of  the  center  of  a  circle  which  is  inscribed 
in  a  triangle  with  a  given  base  and  a  given  vertex  angle  ? 

5.  What  is  the  locus  of  the  mid-point  of  a  line  which  is  drawn 
from  a  given  point  to  a  given  line  ? 

6.  What  is  the  locus  of  the  mid-point  of  a  chord  of  given  length 
in  a  circle  of  given  radius  ? 

7.  What  is  the  locus  of  the  mid-points  of  all  chords  drawn  from 
a  given  point  in  a  given  circle? 

8.  What  is  the  locus  of  the  point  of  intersection  of  the  diagonals 
of  a  rhombus  constructed  on  a  given  line  as  a  base  ? 

9.  What  is  the  locus  of  a  point  equidistant  from  two  given  equal 
circles  V 


176       PLANE  GEOMETRY— BOOK  II 

10.  A  square  (equilateral  triangle,  regular  hexagon)  is  moved 
along  a  straight  line  by  revolving  it  successively  about  its  vertices. 
What  is  the  locus  of  one  vertex  of  the  square  (triangle,  hexagon) 
from  one  point  of  contact  with  the  line  to  its  next  successive  point 
of  contact? 

11.  Two  stations,  A  and  B,  on  the  shore 
of  a  lake  are  900  yds.  apart.    A  ship  C  is 
observed  simultaneously  from  both  stations,      , 
and  the  angles  CAB  and  CBA  are  measured 

by  the  observers  at  each  station.  In  order  to  determine  the  path  of 
the  ship,  additional  measurements  are  made  from  time  to  time  and 
are  telephoned  from  one  station  to  the  other.  Determine  the  path 
of  the  ship  if  the  measurements  are  as  follows : 

/.CAB  /.CBA  /.CAB  /.CBA 

(1)  50°  40°  (2)  71°  69° 

55°  35°  80°  60° 

59°  31°  90°  50° 

70°  20°  100°  40° 

12.  How  may  the  accuracy  of  a  drawing  of  a  circle  be  tested  with 
a  carpenter's  square  ? 

13.  What  locus  problem  is   suggested 
by  a  "  saw-toothed  "  roof  ? 

14.  Three  fortified  islands,  A,  B,  C,  are  so  far  from  land  that  their 
guns  do  not  carry  to  shore.    A  hostile  cruiser  is  reconnoitering  in 
the  vicinity  of  the  islands  and  the  captain  wishes  to  sail  around  them 
by  the  shortest  possible  route,  but  always  out  of  range  of  the  guns. 
Construct  the  course  of  the  cruiser,  using  arbitrary  values  for  the 
ranges  of  the  guns  on  the  islands. 

15.  A  ship,  S)  approaching  land  is  in 
the  vicinity  of  three  prominent  land- 
marks, A,  B,  C.    The  captain  wishes  to 
determine  how  far  he  is  from  the  shore. 
He  finds  that  the  distance  AB  subtends 
at  S  an  angle  of  60°,  while  BC  subtends 

an  angle  of  95°.  From  the  map  the  captain  finds  that  AB  =  16  mi., 
and  BC  =  12  mi.  Explain  how  the  length  of  SB  can  be  found  by 
construction  and  measurement.  (The  "  three-point  problem.") 


BOOK  II 


ITT 


i 


COORDINATES.    SQUARED  PAPER 

297.  Draw  AB±_  to  CD.  From  the  intersection  point  O  lay 
off  equal  segments  on  the  four  rays  OA,  OB,  OC,  OD.  Through 
the  points  of  division  draw  perpendiculars  to  the  lines.  A  net- 
work of  squares  will  be  formed.  Paper  showing  such  an  arrange- 
ment of  squares  is  called  squared  paper.  "  Inch  paper  "  is  ruled 
into  square  inches  and  tenths  of  an  inch ;  "  millimeter  paper  " 
is  ruled  into  square  millimeters  and  square  centimeters. 

The  point  P1    

in  the  figure 
is  represented 
numerically  by 
the  expression 
(15,  10);  this 
means  that  to 
reach  the  point 
Pl  from  0,  the 
pencil's  point 
should  travel  15 
units  to  the  right 
along  OB,  and 
10  units  upward 

along  a  line  parallel  to  OC.  This  process  of  locating  a  point  by 
means  of  two  numbers  is  called  plotting  the  point.  The  point  0  is 
called  the  origin.  The  lines  AB  and  CD  are  called  axes.  The  two 
numbers  are  called  the  coordinates  of  the  point.  The  distance 
from  0  on  AB  is  called  the  abscissa,  and  the  distance  from  the 
line  AB  is  called  the  ordinate  of  the  point.  Distances  measured 
to  the  right  from  0  on  OB  are  considered  as  positive  (+)  num- 
bers, and  distances  to  the  left  from  0  on  OA  are  considered  as 
negative  (— )  numbers.  Distances  vertically  upward  from  A  B  are 
considered  as  positive,  while  distances  vertically  downward  from 
AB  are  considered  negative.  Hence  the  coordinates  of  P2  are 
(-15,10);  those  of  P3  are  (-15,-  10)  ;  those  of  P4are  (15, -10). 


178  PLANE  GEOMETRY— BOOK  II 

EXERCISES 

1.  Locate  on  a  piece  of  squared  paper  four  points,  and  name  their 
coordinates  with  reference  to  two  convenient  axes. 

2.  Locate  the  points  (5,  6),  (-  4,  7),  (-  8,  -  6),  (7,  -  5). 

3.  Plot  the   following  points  and  connect  them   so  as  to  form 
a  pattern :  (0,  0),  (0,  2),  (0,  5),  (0,  7)  ;  (2,  0),  (2,  2),  (2,  5),  (2,  7) ; 
(5,  0),  (5,  2),  (5,  5),  (5,  7)  ;  (7,  0),  (7,  2),  (7,  5),  (7,  7). 

4.  Plot  the  points  (4,  -  5),  (-  4,  -  5),  (0,  8).    What  kind  of  geo- 
metrical figure  is  determined  by  these  points  ? 

5.  What  figure    is    determined    by  the   points   (10,  7),  (6,  —  3), 
(-6, -3),  (-2,7)? 

6.  A  rectangle  drawn  on  squared  paper  is  symmetrical  with  respect 
to  both  axes.    One  vertex  is  (12,  5).    What  are  the  other  vertices? 

7.  Determine,  by  drawing,  the  coordinates  of  the  points  which  are 
5  units  from  the  horizontal  axis,  and  at  a  distance  from  the  origin 
equal  to  13  units. 

8.  Ascertain,  by  drawing,  the  coordinates  of  the  points  which  are 
equidistant  from  the  two  axes,  and  at  a  distance  from  the  origin  equal 
to  10  units. 

298.  The  method  of  coordinates  may  be  conveniently  used  to 
determine  the  locus  of  a  point  when  the  prescribed  condition  in- 
volves the  distances  of  the  point  from  two  perpendicular  lines. 

EXAMPLES 

1.  The  distances  x  and  y  of  a  point  from  two  given  perpen- 
dicular lines  satisfy  the  equation 
(1)  y  =  2*  +  1. 

Determine  the  locus  of  the  point. 

Solution.  We  interpret  x  and  y  as  coordinates  with  respect 
to  the  given  lines  as  axes.  To  locate  a  number  of  points  which 
satisfy  the  given  condition  (1),  assume  values  for  x  and  com- 
pute from  (1)  the  corresponding  values  of  y.  For  example, 
taking  x  =  2,  then  ?/  =  2x2-fl  =  5.  Hence  the  point  (2,  5) 
is  on  the  locus.  In  this  way  we  compute  the  coordinates  of  any 


COORDINATES 


179 


desired  number  of  points  lying  on  the  locus.    The  results  may 
be  conveniently  arranged  in  the  following  table : 


When  x  = 

-3 

-2 

-1 

0 

1 

2 

3 

etc. 

then  y  = 

-5 

-3 

j 

1 

3 

5 

7 

etc. 

ii. 


Now  plot  the  successive 
points  (-3, -5),  (-2, -3), 
(- 1,  - 1),  etc.,  of  this  table. 
All  of  these  points  lie  on 
the  locus,  since  they  satisfy 
the  prescribed  condition. 
They  appear  to  lie  on  a 
straight  line.  Draw  this 
line  LM.  To  prove  that 
this  line  is  the  locus,  that 
is,  to  verify  2  and  3,  §  295, 
would  involve  methods 
which  the  student  woulcj, 
not  now  understand.  It 
must  suffice  here  to  state  the 
fact  that  the  straight  line 
LM  is  the  required  locus. 

In  algebra  the  straight 
line  LM  is  called  the  "  graph  "  of  the  equation  (1). 

Observe  that  (1)  is  an  equation  of  the  first  degree  in  x  and  y. 
It  may  be  shown  that  the  locus  is  always  a  straight  line  if  the 
coordinates  satisfy  an  equation  of  the  first  degree. 

2.  The  sum  of  the  distances  of  a  point  from  two  perpendic- 
ular lines  equals  6.  What  is  the  locus  ? 

Solution.    Taking  the  perpendicular  lines  as  axes  and  the  dis- 
tances as  coordinates,  the  given  condition  may  be  written 
(2)  x  +  y  =  6. 

The  locus  is  now  determined  as  in  Ex.  1,  and  is  a  straight  line. 


180  PLANE  GEOMETRY— BOOK  II 

EXERCISES 

1.  The  distances  x  and  y  of  a  point  from  two  perpendicular  lines 
satisfy  one  of  the  following  equations.    Determine  the  locus  in  each 
case : 

(a)  y  =  3  x  -  1.  (c)   3  x  +  y  =  5.  (e)  4  x  +  3  y  =  12. 

(b)  x  =  2  y  -  6.  (d)  x  -  2  y  =  5.  (f)   2  x  -  3  y  -  6. 

2.  The  difference  of  the  distances  of  a  point  from  two  perpendic- 
ular lines  equals  4.    What  is  the  locus  ? 

3.  What  is  the  locus  of  a  point  whose  abscissa  is  6  units  greater 
than  the  ordinate  ? 

4.  The  distance  of  a  point  from  one  of  two  perpendicular  lines 
equals  twice  its  distance  from  the  other.    Draw  the  locus. 

5.  If  the  coordinates  of  a  point  satisfy  an  equation  which  is  not 
of  the  first  degree,  the  locus  will  usually  be  a  curved  line.    This  is 
the  case  in  the  following  examples.    Draw  the  locus  in  each  case : 

(a)  2  y  =  x2.  (b)  xy  =  3.  (c)  y2  =  4  x. 

6.  If  p  is  the  perimeter  and  a  a  side  of  any  square,  then  p  =  4  s. 
Draw  the  graph  of  this  equation,  plotting  values  of  s  as  abscissas 
and  the  corresponding  values  of  p  as  ordinates. 

7.  The  perimeter  of  an  isosceles  triangle  is  12.    If  x  is  the  base 
and  y  one  of  the  legs,  write  the  equation  satisfied  by  x  and  y  and 
draw  a  graph  of  it. 

8.  The  graphic  method  may  be  used  to  illustrate  geometric  rela- 
tions even  when  the  equation  showing  the  relation  is  not  given.   For 
example,  draw  a  circle  of  radius  2  in.    By  means  of  the  protractor 
mark  off  on  this  circle  from  some  point  A  arcs  of  10°,  20°,  30°,  40°, 
etc.,  to  180°.   Draw  from  A  the  chords  of  these  arcs.  Using  the  num- 
ber of  degrees  in  the  arcs  as  abscissas,  and  the  measured  lengths  of 
corresponding  chords  as  ordinates,  plot  the  points  thus  determined. 
Is  the  graph  a  straight  or  a  curved  line?    Continue  the  graph  by 
increasing  the  number  of  the  arcs  at  intervals  of  10°  up  to  360°. 
What  change  is  there  in  the  corresponding  chords? 

9.  Plot  a  graph  showing  the  relation  between  the  altitude  of  an  equi- 
lateral triangle  and  a  side.    Use  the  bases  of  a  series  of  equilateral 
triangles  as  abscissas,  and  the  corresponding  altitudes  as  ordinates. 


CONSTRUCTIONS  181 

CONSTRUCTION  OF  GEOMETRIC  FIGURES 

299.  The  theorems  of  Book  II  and  the  principles  of  loci 
afford    additional     methods     of    construction     of    geometric 
figures. 

300.  Determining  Parts  of  a  Figure.    A  man  who  wishes  to 
build  a  house  consults  an  architect  with  regard  to  plans.    It  is 
not  necessary  for  the   builder  to  give  the  architect  all  the 
dimensions  of  all  parts  of  the  house.    Certain  measurements 
furnished  by  the  builder  enable  the  architect  to  complete  the 
plans  for  a  building  answering  all  requirements. 

In  like  manner  it  appears  that  only  a  certain  number  of 
parts  of  a  geometric  figure  are  necessary  to  determine  the 
figure.  These  are  known  as  determining  parts. 

TRIANGLES 

301.  The  six  principal  parts  of  a  triangle  are  the  three  sides 
and  the  three  angles. 

How  many  of  these  parts  are  necessary  to  determine  a  triangle  ?  Is 
the  choice  of  parts  otherwise  limited  ? 

302.  In  any  triangle  there  are,  in  addition  to  the  principal 
parts  above  named,  other  lines  and  angles  which  may  serve  to 
determine  the  figure,  such  as  the  medians,  the  altitudes,  the 
bisectors  of  the  interior  angles,  the  angles  between  these  lines, 
and  the  radii  of  the  circumscribed  and  inscribed  circles.    These 
lines  and  angles  are  called  secondary  parts. 

In  the  triangle  ABC  (p.  182)  let  a,  6,  and  c  represent  the  sides  oppo- 
site the  angles  A,  J5,  and  C  respectively. 

Let  raa,  m&,  and  rac  represent  the  medians  on  a,  6,  and  c  respectively. 

Let  ha,  hb,  and  hc  represent  the  altitudes  on  a,  6,  and  c  respectively. 

Let  ta,  tb,  and  tc  represent  the  bisectors  of  the  angles  opposite  a,  6, 
and  c  respectively. 

Let  R  and  r  represent  the  radii  of  the  circumscribed  and  inscribed 
circles  respectively. 


182 


PLANE  GEOMETRY— BOOK  II 


It  will  appear  that,  in  general,  three  parts  are  necessary  and  sufficient 
to  determine  a  triangle,  provided  one  of  these  parts  is  a  line.  But  the 
choice  of  parts  and  of  values  of  these  parts  is  limited  to  some  extent  by 
their  relation  to  one  another. 


303.  Relation  of  Parts.    These  relations  make  necessary  a 
discussion  of  the  limits  within  which  a  solution  of  a  given 
problem  is  possible. 

Try  to  construct  a  triangle  whose  sides  are  5, 8,  and  14  units  respectively. 
Why  is  this  construction  impossible  ? 

Try  to  construct  a  triangle  two  of  whose  angles  are  80°  and  110°. 
Why  is  this  construction  impossible  ? 

How  does  the  altitude  on  one  side  of  a  triangle  compare  in  length  with 
each  of  the  other  two  sides  ?  with  the  corresponding  median  ?  with  the 
bisector  of  the  corresponding  interior  angle  ? 

304.  Method  of  Solution.    When  an  architect  has  had  sub- 
mitted to  him  certain  requirements  for  a  building,  his  practice 
generally  is,  first  to  form  a  mental  image  of  the  structure  as  it 
would  appear  if  completed ;  then  to  make  a  rough  sketch  em- 
bodying his  idea;  then  to  indicate  on  this  sketch  t"he  given 
parts ;  then  to  ascertain  what  other  parts  are  determined  by 
what  is  given  him ;  and  finally  to  make  an  accurate  drawing 
with  the  aid  of  the  information  he  has  thus  obtained. 

305.  A  somewhat  similar  course  of  procedure  is  indicated 
by  the  rule  on  the  opposite  page. 


CONSTRUCTIONS  183 

RULE  FOR  THE  SOLUTION  OF  CONSTRUCTION  PROBLEMS 

1.  Imagine  the  construction  completed,  and  make  a  rough 
sketch  of  the  figure. 

2.  Indicate  on  this  sketch  the  given  parts. 

3.  Ascertain  what  other  parts  are  determined  by  means  of 
the  given  parts. 

4.  If  necessary,  add  such  construction  lines  to  the  rough  sketch 
as  will  make  some  part  of  the  figure  fully  determined. 

5.  Begin  the  actual  construction  of  the  figure  with  that  part 
of  it,  usually  a  triangle,  which  is  fully  determined. 

6.  Complete  the  figure  in  accordance  with  the  general  prin- 
ciples of  construction  lines  (§§  145  seq.). 

7.  Prove  that  the  construction  satisfies  the  given  conditions, 
or  else  show  that  such  proof  is  unnecessary. 

8.  Discuss  the  limits  within  which  the  solution  is  possible, 
and  the  number  of  possible  solutions  when  there  is  more  than 
one,  as  sometimes  happens. 

306.  Determination  of  Triangles.  It  follows  from  the  laws  of 
congruence  of  triangles  that  a  triangle  is  uniquely  determined 
(only  one  solution)  when 

(a)  one  side  and  the  two  adjoining  angles  are  given ; 

(b)  two  sides  and  the  included  angle  are  given ; 

(c)  three  sides  are  given  ; 

(d)  the  triangle  is  a  right  triangle,  and  the  hypotenuse  and 

a  leg  are  given  ; 

(e)  the  triangle  is  a  right  triangle,  and  the  hypotenuse  and 

an  acute  angle  are  given. 

Discuss  the  limits  within  which  each  of  these  constructions  is  possible, 
stating  in  each  case  the  theorem  which  defines  those  limits.  Note  (§  182) 
that  two  angles  of  a  triangle  determine  the  third.  Does  this  fact  make 
possible  an  extension  of  the  foregoing  list  of  conditions  under  which 
a  triangle  is  determined  ?  a  right  triangle  ? 

How  many  principal  parts,  and  what  parts,  determine  an  isosceles 
triangle  ?  an  equilateral  triangle  ? 


184  PLANE  GEOMETRY— BOOK  II 

EXAMPLES 

1.  Construct  the  A  ABC,  having  given  b,  hb,  and  mb. 

Analysis.  Imagine  the  construction  completed,  as  shown  in  the  figure. 
Then  rt.  A  DBE  is  uniquely  determined. 

(Why?)   Also  AE  and  EC   are  determined. 

(Why  ?)  HenceA-4.BC  is  uniquely  determined.     hb 

Construction.  Construct  a  rt.  A  DBE,  using     nib 

mb  as  the  hypotenuse,  and  hb  as  a  leg  (§  163). 
Produce  DE  indefinitely  in  both  directions. 
From  E  lay  off  on  DE  produced  EA  and  EC, 
each  equal  to  one  half  of  b.  Draw  AB  and 
BC.  Proof  is  unnecessary. 

Discussion.  The  construction  is  possible 
only  when  mb>hb.  Otherwise  the  values  of 
the  parts  are  not  limited. 

The  principles  of  loci  also  are  of  assistance  in  determining 
parts  of  a  figure,  when  it  can  be  shown  that  one  or  more  points 
of  the  figure  are  each  the  intersection  of  two  loci  resulting 
from  the  conditions  of  the  problem.  / 

2.  Construct  A-4-BC,  given  /.B,  b,  and  mb. 
Analysis.  Imagine  the  figure  completed.  There 

are  not  sufficient  parts  fully  to  determine  any  one 
of  the  triangles  of  the  figure.  But  it  is  apparent 
that  the  point  B  is  at  the  distance  mb  from  D,  the 
mid-point  of  AC,  and  also  that  it  is  the  vertex 
of  an  angle  equal  to  the  given  ZB,  the  sides  of 
which  pass  through  A  and  C. 

Hence  one  locus  of  the  point  B  is  a  circle  about 
D  as  a  center,  with  a  radius  equal  to  mb ;  and  the 
other  locus  of  the  point  B  is  an  arc  upon  A  C  as  a 
chord,  containi ng  Z. B  as  an  inscribed  angl e  (§  293) ; 
and  B  is  determined  as  either  of  the  two  points 
of  intersection  of  the  two  loci. 

Construction.  Layoff  A  C  =  b.  At  C  construct  ZACE  =  ZB.  At  C  draw 
CF±toCE.  Bisect  ACinD.  At  D  draw  DG±  to  AC,  intersecting  CF  in  0. 
About  0  as  a  center,  with  a  radius  equal  to  OC,  describe  a  circle.  About 
D  as  a  center,  with  a  radius  equal  to  mb,  describe  another  circle  cutting 
the  first  circle  in  B.  Draw  BA  and  BC.  A  ABC  is  the  triangle  required. 
(Proof  to  be  completed.) 

Discussion.  How  many  solutions  are  possible  ?  (Why  ?)  When  is  the 
solution  impossible  ? 


CONSTRUCTIONS  185 

EXERCISES 
TRIANGLES 

1.  Construct  a  triangle,  having  given  two  sides  and  the  angle 
opposite  one  of  them  (a,  b,  A  A}. 

Solution.  Draw  a  line  A  C  equal  to  b.  At  A  construct  Z  CA  E  equal 
to  the  given  /.A.  About  C  as  a  center,  and  with  a  radius  equal  to 
the  line  a,  describe  an  arc  intersecting  AE  in  B. 

Discussion.  If  /.  A  is  a  right  angle,  how  must  a  compare  in  length 
with  6  ?  (Why  ?) 

If  Z  A  is  an  obtuse  angle,  how  must  a  compare  in  length  with  6? 
(Why?) 

If  Z.  A  is  an  acute  angle,  has  the  side  a  any  lower  limit  of  length  ? 
Under  what  conditions  are  there  two  solutions  of  the  problem  ? 

Construct  an  equilateral  triangle,  having  given : 

2.  The  altitude. 

3.  The  sum  of  the  altitude  and  one  side.     (Construct  in  the 
figure  for  analysis  a  line  representing  the  given  sum.) 

4.  The  radius  of  the  circumscribed  circle. 

5.  The  radius  of  the  inscribed  circle. 

Construct  a  right  triangle,  having  given  : 

6.  A  leg  and  the  opposite  acute  angle. 

7.  A  leg  and  the  altitude  on  the  hypotenuse. 

8.  A  leg  and  the  median  on  the  other  leg. 

9.  An  acute  angle  and  the  altitude  on  the  hypotenuse. 

10.  A  leg  and  the  bisector  of  the  right  angle.  (How  many  degrees 
in  the  angle  formed  by  the  two  given  lines  ?) 

11.  The  hypotenuse  and  the  sum  of  the  legs. 

12.  The  hypotenuse  and  the  difference  of  the  legs. 

13.  An  acute  angle  and  the  sum  of  the  legs. 

14.  One  leg  and  the  radius  of  the  inscribed  circle. 

15.  The  hypotenuse  and  the  radius  of  the  inscribed  circle. 

16.  The  radii  of  the  inscribed  and  circumscribed  circles. 


186  PLANE  GEOMETRY— BOOK  II 

Construct  an  isosceles  triangle,  having  given  : 

17.  The  base  and  the  altitude. 

18.  The  base  and  the  vertex  angle. 

19.  The  base  and  the  altitude  on  one  of  the  legs. 

20.  The  vertex  angle  and  the  altitude  upon  the  base. 

21.  A  base  angle  and  the  sum  of  the  base  and  one  leg. 

Construct  &ABC  (a,  b,  c),  having  given: 

22.  a,  I,  V  28.  a,  ha,  Tic.  34.  a,  mb,  Z  C. 

23.  a,  b.  mb.  29.  a,  ha,  /.B.  35.  a  +  b,  c,  Z  B. 

24.  a,  b,  hc.  30.  «,  ha,  /.A.  36.  a  +  b  + c,ZA,ZC. 

25.  a,  ma,  ha.  31.  ha,  Z  B,  Z  C.  37.  a  +  b,  c,  ha. 

26.  «,  hb,  ZA.  32.  ha,  hc,  Z  C.  38.  a,  mb,  mc. 

27.  a,  ma,  Z  C.  33.  a,  w6,  c.  39.  ma,  mfc,  wic. 

40.  Mention  three  groups  of  three  parts  each  which  do  not  afford 
a  solution  of  the  triangle,  and  give  reasons  why  they  do  not. 

QUADRILATERALS 

(Among  the  "  secondary  parts  "  of  a  quadrilateral  are  the  diagonals  ; 
of  a  parallelogram,  the  diagonals  and  the  altitudes ;  of  a  trapezoid, 
the  altitude  and  the  mid-line.) 

41.  How  many  parts  are  necessary  to  determine  a  square?    What 
parts? 

42.  How   many  parts  are  necessary  to   determine  a  rhombus? 
What  parts? 

43.  How  many  parts  are  necessary  to  determine  a  rectangle  ?  a 
parallelogram  ?  a  trapezoid  ?  a  quadrilateral  in  general  ? 

44.  Construct  a  square,  having  given  the  sum  of  the  diagonal  and 
one  side. 

Construct  a  rhombus,  having  given : 

45.  The  diagonals. 

46.  One  angle  and  one  of  the  diagonals. 

47.  The  base  and  the  altitude. 


CONSTRUCTIONS  187 

Construct  a  rectangle,  having  given : 

48.  One  side  and  a  diagonal. 

49.  The  perimeter  and  a  diagonal. 

50.  One  side  and  the  corresponding  angle  between  the  diagonals. 
Construct  a  trapezoid,  having  given : 

51.  The  four  sides. 

52.  The  bases  and  the  diagonals. 

53.  The  bases  and  the  base  angles. 
Construct  a  parallelogram,  having  given : 

54.  Two  sides  and  the  altitude  on  one  of  them. 

55.  Two  diagonals  and  one  side. 

56.  Two  diagonals  and  the  included  angle. 
Construct  an  isosceles  trapezoid,  having  given : 

57.  The  bases  and  one  diagonal. 

58.  The  bases  and  the  altitude. 

59.  The  bases  and  one  base  angle. 

Construct  a  trapezium,  having  given : 

60.  Three  sides  and  the  two  included  angles. 

61.  The  segments  of  the  diagonals  made  by  their  point  of  inter- 
section, and  the  angle  between  the  diagonals. 

Construct  a  circle,  having  given : 

62.  That  it  has  a  radius  r  and  is  tangent  to  two  intersecting  lines. 

63.  That  it  has  a  radius  r  and  is  tangent  to  a  given  line  and  also 
to  a  given  circle. 

64.  That  it  touches  two  given  parallel  lines  and  passes  through 
a  given  point. 

65.  That  it  touches  a  given  line  at  a  given  point,  and  passes 
through  a  given  point  without  that  line. 

66.  That  it  has  a  radius  r,  passes  through  a  given  point,  and 
touches  a  given  circle. 

67.  Construct  the  bisector  of  the  angle  which  two  lines  would  form 
if  produced,  without  actually  producing  them  to  their  intersection. 

68.  In  a  given  triangle,  to  join  two  sides  by  a  line  of  a  given 
length,  which  shall  be  parallel  to  the  base. 


188 


PLANE  GEOMETKY— BOOK  II 


REVIEW  EXERCISES 

1.  If  two  intersecting  chords  make  equal  angles  with  the  diameter 
drawn  through  their  point  of  intersection,  the  chords  are  equal. 

(Let  fall  Js  upon  the  chords  from  the  center.) 

2.  The  mid-line  of  a  trapezoid  circumscribed  about  a  circle  is 
equal  to  one  fourth  the  perimeter  of  the  trapezoid. 

3.  If    from  any   point  on  a   circle   a   chord  and  a  tangent  are 
drawn,  the  perpendiculars  let  fall  upon  them  from  the  mid-point  of 
the  intercepted  arc  are  equal. 

(Join  the  mid-point  of  the  arc  to  the  given  point  on  the  circle.) 

4.  Two  circles  cut  each  other  in  two  points 
M  and  N.    Through  M  the  line  PQ  is  drawn 
meeting  the  circles  in  P  and  Q.    Prove  that  the 
ZPNQ  is  constant  for  all  positions  of  PQ,  so 
long  as  it  passes  through  M. 

5.  Two    circles    touch    each    other    in    P. 
Through  this  point  of  tangency  two  lines  are 
drawn,   one  meeting  the  circles  in  M  and  N, 
the  other  in  Q  and  R,  respectively.    Prove  that 
MQ  II  NR. 

6.  Two  circles  touch  each  other  in  P.    AB 
is  a  line  through  P  meeting  the  circles  in  A 
and  B.    Prove  that  the  tangents  at  A  and  B  are 
parallel. 

7.  Two  circles  touch  each  other  internally  at  P. 
MN  is  a  chord  of  the  larger,  tangent  to  the  smaller 
at  C.    Prove  Z.MPC  =  /.  CPN. 

8.  If  through  the  points  of  intersection  of  two 
circles  parallels  are  drawn  terminating  in  the  circles, 
these  parallels  are  equal. 

9.  If  two  circles  are  tangent  to  each  other  externally,  (a)  the 
common  internal  tangent  bisects  the   common   external   tangent ; 
(b)  the  tangents  to  the   circles    from   any  point   of  the  common 
internal  tangent  are  equal. 

State  (b)  as  a  locus  problem. 


EXEKCISES 


189 


10.  When  two  circles  are  tangent,  internally  or  externally,  the 
point  of  tangency  may  be  regarded  as  a  transition  point  from  one 


FIG.  1 


circle  to  the  other.  Arcs  of  the  two  circles  which  meet  at  the  point 
of  tangency  then  seem 
to  form  one  continuous 
curve.  This  principle  is 
applied  in  the  construc- 
tion of  moldings,  spi- 
rals, and  other  decorative 
forms,  and  in  laying 
out  curves  in  railroad 
building. 

Copy  each  of  the  sim- 
ple moldings  in  Fig.  1.  In 
(#)  the  radius  of  the  up- 
per quadrant  is  one  third  the 
height  of  the  molding.  In  (&) 
and  (7)  the  arcs  are  each  60°. 

11.  Copy  on  a  larger  scale 
each  of  the  composite  mold- 
ings in  Fig.  2,  and  explain 
the  construction. 

12.  Draw  a  figure   illus- 
trating the  principle  of  tan- 
gency of  circles  as  shown  in 
the  cross  section  of  a  ball- 
bearing   wheel     and     axle. 
What    advantage    has   this 
bearing  over   the    ordinary 

bearing?  FIG.  2 


190 


PLANE  GEOMETRY— BOOK  II 


D 


13.  A  form  which  appears  frequently  in  architectural  design  is 
the  equilateral  Gothic  arch,  which  is  formed  by  using  two  sides  of 
the  equilateral  triangle  as  chords  and  the  third  side 

as  a  radius,  and  drawing  two  arcs,  as  shown  in  the 
figure. 

It  is  required  to  inscribe  a  circle  in  such  an  arch. 
Describe  a  method  of  construction  based  on  the  fol- 
lowing analysis : 

If  0  is  the  center  of  the  required  circle, 
then  FB  passes  through  0,  and  OC  —  OF. 
(Why  ?)  Whence  A  OCB  =  A  OFE  (a.  s.  a.) 
and  EF=  CB.  Whence  also  A  BEC  =  A  BEF 
(rt.  A  h.  1.).  Hence  CE  =  BF  =  A B. 

14.  In   window  designing,    several   equi- 
lateral  Gothic    arches   are  often  combined 
with  one  or  more  circles.  A  simple  combina- 
tion of  this  kind  is  shown  in  the  adjoining 
figure.  Copy  this  figure  and  explain  its  mode 
of  construction. 

(The  radius  to  the  point  of  common  tan- 
gency  L  passes  through  H,  the  center  of  the 
circle,  and  through  M,  the  point  of  common 
tangency  of  the  circle  and  one  of  the  smaller 
arcs.  It  follows  that  AH  equals  three  fourths 
oiAB.  Why?) 


D 


15.  An  opaque  body  intercepts  the  rays  of  light  which  shine 
upon  it,  leaving  a  dark  space  behind.  This  space  is  called  the 
shadow.  The  shadow  will  appear  dark  in  cross  section  on  a  screen 
if  the  screen  is  placed  beyond  the 
opaque  body  opposite  to  the  source 
of  light. 

When  the  source  of  light  is  a  very 
small  body,  as,  for  example,  the  lumi- 
nous point  of  the  carbon  of  an  electric  arc  light,  it  may  be  regarded 
as  a  geometrical  point.  In  that  case  the  edge  of  the  shadow  as  shown 
on  the  screen  is  clear  and  distinct.  Suppose  that  the  opaque  body  is 
a  sphere.  Imagine  a  plane  passed  through  the  source  of  light  and 


EXERCISES 


191 


the  center  of  the  sphere.  The  boundary  of  the  shadow  in  this  plane 
is  then  determined  by  two  lines  of  indefinite  length  drawn  from  the 
source  of  light,  tangent  to  the  circle. 

If  the  source  of  light  is  comparatively  large,  the  shadow  will  be 
made  up  of  two  portions  :  the  umbra,  that  is,  the  portion  of  space  from 
which  all  light  from  the  source  is  ex- 
cluded ;  and  the  penumbra,  which  is 
the  portion  of  space  from  which  part 
of  the  light  is  excluded.  Explain 
how  the  boundaries  of  the  umbra 
and  penumbra  may  be  determined. 

(a)  How  are  the  umbra  and  penumbra  affected  by  a  change  in 
the  distance  apart  of  the  luminous  and  opaque  bodies? 

(b)  If  a  croquet  ball  is  held  in  the  path  of  the  sun's  rays,  is  there 
any  penumbra  ?    Explain. 

(c)  What  may  be  said  of  the  umbra  if  the  luminous  sphere  and 
the  opaque  sphere  are  of  the  same  size  ? 

(d)  What  is  the  shape  of  the  umbra  if  the  luminous  sphere  is 
larger  than  the  opaque  sphere,  as  in  the  case  of  the  sun  and  the  earth  ? 

(e)  When  the  moon  passes  through 
the  earth's  shadow  a  "  lunar  eclipse  " 
occurs.     The    eclipse    is    "  total "    or 
"  partial "  according  as  the  moon  is 
entirely  or  partly  in  the  umbra  (see 

the  above  figure).  A  solar  eclipse  arises  if  the  moon's  shadow  passes 
across  the  surface  of  the  earth.  Sometimes  the  earth  does  not  enter 
the  moon's  umbra  at  all  on  such  an  occasion.  Discuss  the  relative 
positions  of  the  sun,  the  moon,  and  the  earth  in  that  case. 

16.  If  two  circles  intersect  and  a  line  is  drawn  through  each  point 
of  intersection  terminated  by  the  circumferences,  the  chords  joining 
the  ends  of  these  lines  are  parallel. 

17.  Prepare  a  summary  of  the  properties  of  two  circles  in  each  of 
the  six  possible  positions. 

18.  What  methods  are  furnished  by  the  propositions  of  Book  II  for 
proving  the  equality  of  line-segments?  of  angles?  of  arcs?  of  chords? 

19.  Prepare   a   complete   list  of  all  applied   problems  given  in 
Book  II,  and  also  a  list  of  the  principles  of  loci. 


192  PLANE  GEOMETRY— BOOK  II 

LOCUS  PROBLEMS 

1.  In  the  rectangle  ABCD  the  side  AB  is  twice  as  long  as  the 
side  BC.    A  point  E  is  taken  on  the  side  AB,  and  a  circle  is  drawn 
through  the  points  C,  D,  and  E.    Construct  the  path  of  the  center 
of  the  circle  as  E  moves  from  A  to  B. 

2.  Given  a  square  with  each  side  3  in.  long.    Construct  the  locus 
of  a  point  P  such  that  the  distance  from  P  to  the  nearest  point  of 
the  square  is  1  in. 

3.  A  straight  line  3  in.  long  moves  with  its  extremities  on  the 
perimeter  of  a  square  whose  sides  are  4  in.  long.    Construct  the  locus 
of  the  middle  point  of  the  moving  line. 

4.  A  circular  basin  16  in.  in  diameter  is  full  of  water,  and  upon 
the  surface  there  floats  a  thin  straight  stick  1  ft.  long.    Shade  that 
region  of  the  surface  which  is  inaccessible  to  the  middle  point  of  the 
stick  and  describe  accurately  its  boundary. 

5.  The  image  of  a  point  in  a  mirror  is  apparently  as  far  behind 
the  mirror  as  the  point  itself  is  in  front.    If  a  mirror  revolves  about 
a  vertical  axis,  what  will  be  the  locus  of  the  apparent  image  of  a 
fixed  point  1  ft.  from  the  axis  ? 

6.  Upon  a  given  base  is  constructed  a  triangle,  one  of  whose 
base  angles  is  double  the  other.    The  bisector  of  the  larger  base 
angle  meets  the  opposite  side  at  the  point  P.    Find  the  locus  of  P. 

7.  Find  the  locus  of  the  middle  points  of  straight  lines  drawn 
between  two  parallel  lines. 

8.  Find  the  locus  of  the  extremity  of  a  tangent  of  given  length 
drawn  to  a  given  circle. 

9.  Find  the  locus  of  the  center  of  a  circle  which  has  a  given 
radius  and  is  tangent  to  a  given  circle. 

10.  Find  the  locus  of  the  points  of  contact  of  tangents  drawn 
from  a  given  point  to  a  given  set  of  concentric  circles. 

11.  Find  the  locus  of  the  centers  of  circles  which  touch  two  given 
concentric  circles. 

12.  An  angle  of  60°  moves  so  that  both  of  its  sides  touch  a  fixed 
circle  of  radius  5  ft.    What  is  the  locus  of  the  vertex  ? 

13.  If  through  a  fixed  point  within  a  circle  a  chord  is  drawn,  find 
the  locus  of  the  middle  point  of  that  chord. 


BOOK  HI 

AREA 
PRELIMINARY  DEFINITIONS  AND  EXERCISES 

307.  Two  geometric  magnitudes  are  said  to  be  equal  if  they 
are  of  the  same  size. 


F 


Thus,  the  expression  F  =  F'  signifies  that  the  quadrilateral  F  and 
the  triangle  F'  occupy  equal  portions  of  the  plane. 

308.  It  follows  at  once  that  congruent  figures  are  also  equal. 
Also,  if  two  figures  are  composed  of  parts  that  are  respectively 
congruent,  they  are  evidently  equal,  though  not  themselves 
necessarily  congruent. 


For  example,  two  congruent  triangles  may  be  placed  so  as  to  form 
either  a  parallelogram  or  a  kite.  The  two  resulting  figures  are  equal, 
but  not  congruent. 

309.  To  transform  a  figure  means  to  construct  another  figure 
of  different  form,  equal  to  it.. 

Thus,  we  shall  learn  how  to  transform  any  polygon  into  a  tri- 
angle, any  triangle  into  a  square,  etc.  The  kite  in  the  above  figure 
is  a  transformation  of  the  parallelogram. 

310.  To  find  the  size  of  any  plane  figure,  it  is  necessary  to 
select  some  unit  of  surface,  and  to  ascertain  the  number  of  times 
this  unit  is  contained  in  the  given  figure.  A  square,  constructed 

193 


194 


PLANE  GEOMETRY— BOOK  III 


on  a  side  of  convenient  length,  is  the  most  satisfactory  unit 
for  practical  purposes.  The  standard  square  units  are  the 
square  centimeter  (sq.  cm.),  the  square  meter  (sq.  m.),  the 
square  inch  (sq.  in.),  the.  square  foot  (sq.  ft.),  and  other 
squares,  each  of  whose  sides  is  a  standard  linear  unit. 

311.  The  area  of  a  figure  is  the  number  showing  how  many 
square  units  of  a  given  kind  it  contains. 

312.  The  axioms  of  equality  and  inequality  (§§132,  222) 
apply  to  areas  as  well  as  to  line-segments  and  angles. 


E 


EXERCISES 

1.  Fold  a  rectangular  sheet  of  paper  A  BCD  into  two  equal  parts, 
by  placing  EC  on  AD,  etc.    By  repeating 

this  process,  divide  the  given  rectangle  into 
small  rectangles. 

2.  Repeat  Ex.  1,  using  a  square  piece  of 
paper  instead  of  the  rectangle.  What  form 
will  each  of  the  small  rectangles  take  ? 

3.  If  in  Ex.  1  the  small  rectangle  were  taken  as 
the  unit  of  area,  what  would  be  the  area  of  the  large 
rectangle  ? 

4.  In  the  second  figure  the  area  of  the  square  is 
said  to  be  16  square  nnits.    What  does  that  mean  ? 

5.  Draw  a  rectangle  whose  sides  are  7  cm.  and  5  cm.    Divide  it 
into  squares.    Show  that  it  contains  35  sq.  cm. 

313.  Summary.    From  the  foregoing  exercises  we  conclude  : 

1.  The  area  of  a  rectangle  whose  sides  contain  a  units  and 
b  units  of  the  same  kind  respectively  is  a  x  b  square  units. 

2.  The  area  of  a  square  whose  side  contains  a  units  is  a? 
square  units. 

NOTE.  The  base  and  the  altitude  of  a  rectangle  are  often  called  its 
dimensions.  In  propositions  relating  to  areas  the  words  rectangle,  tri- 
angle, etc.,  are  often  used  for  area  of  rectangle,  area  of  triangle,  etc. 


I 

mas       D 
large 

re  is        A 

f 

AEEAS 


195 


EXERCISES 

1.  In  each  of  the  following  exercises  plot  the  points  (see  §  297), 
join  them  by  straight  lines  in  the  order  given,  and  determine  the 
number  of  square  units  in  the  figure  formed : 

(a)  (4,2),  (-4,2),  (-4,  -5),  (4,  -5). 

(b)  (3,  2),  (-  1,  2),  (-  1,  -  5),  (3,  -  5). 

(c)  (0,  6),  (4,  6),  (4,  -  3),  (0,  -  3). 

(d)  (-  2,  3),  (-  2,  9),  (-  6,  9),  (-  6,  3). 

2.  If  a  and  b  represent  the  number  of  units  in  the 
legs  of  a  right  triangle  ABC,  show  that  the  triangle 

7 

contains  ^— —  square  units. 

3.  The    annexed   diagrams   illustrate 
how  the  area  of  an  irregular  plotted  figure 
may  be  found.    When  one  side  of  the 
figure  coincides  with  a  line  of  the  squared 
paper,  the  figure  may  easily  be  divided 
into  rectangles  and  right  triangles. 

When  that  is  not  the  case,  lines  may 
be  drawn  through  the  vertices  of  the 
figure,  parallel  to  the  ruled  lines,  thus 
forming  a  circumscribed  rectangle.  By 
subtracting  from  that  rectangle  certain 
triangles,  the  required  area  is  obtained. 

In  each  of  the  following  exercises  plot  the  points  as  in  Ex.  1,  and 
find  the  areas  of  the  plotted  figures  by  the  method  just  outlined  : 

(a)  (1,  1),  (10,  1),  (7,  7),  (3,  5). 

(b)  (4,  2),  (2,  4),   (-  2,  4),  (-  4,  2),  (-  4,  -  2), 
(-  2,  -  4),  (2,  -  4),  (4,  -  2). 

(c)  (0,  6),  (6,  0),  (0,  -  6),  (-  6,  0). 

(d)  (0,  3),  (4,  0),  (6,  4),  (2,  7). 

4.  The  accompanying  figure  illustrates  the  formula : 
(a  +  &)2  =  a2  +  2  ab  +  b2.   Explain. 

5.  Show  by  a  diagram  that  the  square  on  ^  cm.  =  \  sq.cm. 

(1        \2      1  /I        \2       1 

-cm.)  = -sq.cm.,  and  that  i- cm.  1    =  —  sq.cm. 
o         /        9  \n        I        n 


196 


PLANE  GEOMETEY— BOOK  III 


6.  Construct  a  square  on  2|  cm.    Prove  geometrically  that  the 
area  is  6^  sq.  cm.    Verify  numerically. 

Suggestion.  (2£)2  =  (2  +  £)2  =  4  +  2  +  £  =  6i. 

7.  Construct  a  rectangle  whose  sides  are  2^  in.  and  1^  in.  re- 
spectively.   Show  geometrically  that  its  area  is  (2^  x  1^)  sq.  in.  = 
3  J  sq.  in. 

Suggestion.  2J-  x  1J  =  f  X  j  =  V  •  f  =  *££-  =  3 J. 

8.  The  sides  of  a  rectangle  are  2.4  cm.  and  3.1  cm.    Prove  that 
its  area  is  (2.4  x  3.1)  sq.  cm. 

Suggestion.      2.4   cm.  =  24    mm.      Hence,    24   mm.  x  31   mm.  = 
744  sq.  mm.  =  7.44  sq.  cm.  =  (2.4  x  3.1)  sq.  cm. 

9.  If  the  area  of  the  square  A  BCD  is  1  sq.  in.,  the  outer  square 
must  contain  2  sq.  in.  (Why  ?)  Hence,  if  the 

rules  developed  above  are  to  apply  to  this 
square,  we  must  assume  that  its  side  LM 
is  V2  in.  long.  But  the  value  of  V2  is 
1.414- •«.  This  is  a  decimal  which  can 
never  be  completely  written.  Does  this  con- 
vince you  that  there  are  lines  of  definite 
length  which  we  can  measure  only  approxi- 
mately in  terms  of  a  given  unit  ?  Draw  the 
figure  on  squared  paper. 

10.  If   the   dimensions  of   a   rectangle  are  V2  =  1.414  -  •  •,  and 
V3  =  1.732  •  •  •,  we  shall  assume  that  its  area  is  V2  •  V3  =  V6  = 
2.449  •  •  •  sq.  ft.    Continue  the  following  table  and  show  that  the 
products   constantly   approach    the   exact    area    of    the    rectangle ; 

that  is,  V6  sq.  ft. 

1.4  x  1.7  =  2.38. 

1.41  x  1.73  =  ..-. 
1.414  x  1.732 

11.  Find  the  area  of  a  rectangle  if  the  sides  a  and  b  have  the 
values  indicated  in  the  following  table : 


D 


a 

I 

II 

Ill              IV 

V 

VI 

VII 

7  cm. 

.56  in. 

Mt. 

89  mm. 

V3 

V7 

z  +  y 

b 

6.5  cm. 

.73  in. 

6.5ft. 

4.7cm. 

V5 

V2 

x-y 

KATIO  197 

314.  Area  of  a  Rectangle.    From  the  foregoing  exercises  it 
appears  that  the  area  of  a  rectangle  in  square  units  is  equal 
to  the  product  of  its  two  dimensions  in  linear  units,  even  if  those 
dimensions  are  fractional,  decimal,  or  irrational. 

RATIO 

315.  If  two  magnitudes  of  the  same  kind,  such  as  two  line- 
segments,  contain  a  certain  unit  a  and  b  times  respectively, 

then  the  quotient  -  is   often  called  the  ratio  of  these  two 
magnitudes. 

For  example,  the  ratio  of  a  line  10  ft.  long  to  one  30  ft.  long  is  i$,  or  J. 
The  ratio  of  m  to  n  is  often  written  ra :  n,  or  m  -±-  n. 

In  the  ratio  a  :  6,  a  and  6  are  called  the  terms  of  the  ratio  ;  a  is  called 
the  antecedent  and  b  the  consequent. 

316.  Since  ratios  are  really  fractions,  their  properties  are 
the  properties  of  fractions.    Hence 

The  value  of  a  ratio  is  not  changed  if  both  of  its  terms  are 
multiplied  or  divided  by  the  same  number. 

317.  When  two  ratios  a  :  b  and  c  :  d  are  equal,  the  four  num- 
bers a,  b,  c,  d,  are  said  to  be  in  proportion  or  to  be  proportional. 
This  equality  of  ratios  may  be  written  in  any  one  of   the 
following  forms : 

-r  =  ->     a:b  =  c  :  d,     a  :  b  :  :  c  :  d. 
b       d 

These  are  read  "  a  is  to  b  as  c  is  to  d." 

318.  Since  the  ratio  between  two  magnitudes  of  the  same 
kind  is  obtained  exactly  or  approximately  as  the  quotient  of 
their  numerical  measures,  it  is  customary  to  extend  the  use 
of  the  term  "  ratio  "  to  include  the  quotient  of  the  numerical 
measures  of  two  magnitudes  of  different  kinds. 

Thus  the  number  of  pounds  of  pressure  per  square  foot  of  area  of  a 
gas  inclosed  in  a  vessel  is  defined  as  the  ratio  of  the  total  pressure  to  the 
total  area  of  the  inclosing  vessel. 


198  PLANE  GEOMETRY— BOOK  III 

319.  The  ratio  of  two  magnitudes  of  the  same  kind  is  inde- 
pendent of  the  unit  by  which  the  magnitudes  are  measured, 
since  a  change  in  the  unit  results  merely  in  multiplying  or 
dividing  the  two  terms  of  the  ratio  by  the  same  number. 

Thus  the  ratio  of  the' areas  of  two  rectangles  is  the  same,  whether  they 
are  both  measured  in  square  inches  or  in  square  centimeters. 

320.  The  ratio  of  two  magnitudes  of  different  kinds  is  de- 
pendent upon  the  units  by  which  the  two  magnitudes  are 
measured. 

Thus  the  ratio  of  the  weight  of  a  physical  solid  in  pounds  to  its  volume 
in  cubic  feet  is  not  the  same  as  the  ratio  of  its  weight  in  grams  to  its 
volume  in  cubic  centimeters. 

EXERCISES 

1.  Simplify  the  following  ratios  :  8  :  24  ;  3^:7;  4^:4^;  .4  :  .03  ; 
(a  +  ft)2  :  (a  +  ft). 

2.  What  is  the  ratio  of  a  straight  angle  to  a  right  angle  ?  of  the 
interior  angle  of  a  regular  hexagon  to  the  interior  angle  of  an  equi- 
lateral triangle  ? 

3.  Divide    an    angle    of    180°   into   five   parts   in   the  ratio   of 
1:2:3:4:5.    How  many  degrees  in  each  angle  ? 

4.  Divide  50  in  the  ratio  of  m  :  n. 

5.  The  sides  of  a  triangle  are  in  the  ratio  of  3  :  5  : 7.  The  perim- 
eter of  the  triangle  is  30.    Find  the  length  of  each  side. 

6.  Find  the  ratio  of  the  areas  of  two  rectangles  if  their  respective 
dimensions  in  feet  are  12  by  27  and  18  by  20  ;  3£  by  7^  and  2§  by  9. 

7.  What  is  the  ratio  of  the  area  of  a  given  square  to  the  area  of 
the  square  on  its  diagonal  ? 

8.  What  is  the  ratio  of  two  rectangles,  the  first  of  base  10  and 
altitude  x,  and  the  second  of  base  12  and  altitude  #? 

9.  What  is  the  ratio  of  two  rectangles  on  the  same  base  ft,  if 
their  respective  altitudes  are  15  and  20  ? 

10.  What  is  the  ratio  of  the  area  in  square  feet  of  a  square,  each 
of  whose  sides  is  6  feet,  to  the  length  of  one  of  its  sides?  Is  the  ratio 
the  same  if  the  measurements  are  taken  in  inches  ?  in  yards  ? 


AREAS 


199 


AEEAS  OF  SIMPLE, FIGURES 

321.  Fundamental  Principle.    The  area  of  a  rectangle  is  equal 
to  the  product  of  its  base  and  altitude. 


Thus  if  a  and  b  are  the  altitude  and  base  respectively  of  the  rectangle 
whose  area  is  E,  then       R  =  ab  gquare  unitg> 

322.  COROLLARY  1.   Two  rectangles  are  to  each  other  as  the 
products  of  their  bases  and  altitudes. 

For  if  R  =  «6,  and  It'  =  a'b',  then  —  = 

.K'      cn'b' 

323.  COROLLARY  2.   Two  rectangles  having  equal  bases  are 
to  each  other  as  their  altitudes. 

324.  COROLLARY  3.   Two  rectangles   having  equal  altitudes 
are  to  each  other  as  their  bases. 

325.  COROLLARY  4.   Tivo  rectangles  having  equal  altitudes 
and  equal  bases  are  equal. 

The  above  corollaries  may  be  written  symbolically  as  follows : 


326.  COROLLARY  5.   The  area  of  a  right  triangle  is  equal  to 
one  half  the  product  of  its  legs. 

°\ 
h 


327.  COROLLARY  6.   The  area  of  a  kite  (rhombus,  square,  see 
151)  is  equal  to  one  half  the  product  of  its  diagonals. 


200 


PLANE  GEOMETRY— BOOK  III 


EXERCISES 

1.  The  edge  of  a  cube  is  4  in.  How  many  square  feet  in  its  entire 
exterior  surface? 

2.  A  map  is  drawn  so  that  1  cm.  represents  1000  m. 
What  area  is  represented  by  1  sq.  cm.  ? 

3.  The  dimensions  of  a  rectangular  window  are  5  ft. 
and  3  ft.    It  is  to  be  divided  into  rectangles  and  squares 
as  shown  in  the  figure.    Determine  the  dimensions  of 
these  parts. 

4.  The  accompanying  diagrams  represent  cross  sections  of  steel 
beams.    Determine  the  areas  of  the  cross  sections,  using  the  dimen- 
sions given  in  millimeters  in  the  following  table  : 


I 

II 

III 

IV 

b 

96 

72 

112 

128 

w 

12 

9 

14 

16 

h 

192 

144 

250 

300 

t 

8 

6 

10 

10 

h- &— * 


5.  The  diagonals  of  a  square  are  each  10  ft.  long.    Find  the  area 
of  the  square  (§  327). 

6.  The  diagonals  of  a  rhombus  are  8  ft.  and  7  ft.  long.    Find  the 
area  of  the  rhombus  (§  327). 

7.  If  p  and  A  represent  the  perimeter  and  the  area  respectively 
of  a  rectangle,  determine  the  dimensions,  x  and  y,  in  the  exercises 
shown  in  the  following  table  : 


I 

II 

III 

IV 

P 

12 

16 

30 

2a  +  26 

A 

8 

15 

56 

ab 

8.  The  Greek  historian  Thucydides  (430  B.C.)  estimated  the  size 
of  the  island  of  Sicily  by  means  of  the  time  it  took  to  sail  around  it. 
What  was  the  fallacy  in  his  method? 


AREAS 
PROPOSITION  I.    THEOREM 


201 


328.   The  area  of  a  parallelogram  is  equal  to  the  product  of 
its  base  and  altitude. 


E     C 


Given  the  parallelogram  ABCD,  with  its  base  AB  equal  to  &,  and 
its  altitude  BE  equal  to  h. 

To  prove  that  the  area  of  the  HA  BCD  =  b  x  h. 

Proof.    1.  Draw  the  line  A  F II  to  BE,  meeting  CD  produced  at  F. 

Then  A BEF  is  a  rectangle.  Why? 

2.  Also  EH  ABC D  and  rectangle  ABEF  have  the  same  base 
and  the  same  altitude.  Why  ? 

3.  But  AADF=ABCE.  Why? 

4.  Then  area  ABED  +  ABCE  =  area  ABED  +  AADF.   Ax.  2 
That  is,  HABCD  =  rectangle  ABEF. 

5.  But  area  rectangle  ABEF  =  bh.  Why? 

.'.  area  O  ABCD  =  bh. 

329.  COROLLARY  1.   Two  parallelograms  are  to  each  other  as 
the  products  of  their  bases  and  altitudes. 

330.  COROLLARY  2.   Two  parallelograms  having  equal  bases 
are  to  each  other  as  their  altitudes. 

331.  COROLLARY  3.    Two  parallelograms  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 

332.  COROLLARY  4.   Two  parallelograms  having  equal  bases 
and  equal  altitudes  are  equal. 


202  PLANE  GEOMETRY— BOOK  III 

EXERCISES 

1.  If  P  and  P'  are  the  areas  of  two  parallelograms  with  bases 
ft  and  ft',  and  altitudes  h  and  h',  interpret  the  following  equations, 
giving  a  statement  in  full  for  each  : 
P       bh 


2.  Show  that  the  area  of  a  parallelogram  may  be  independent  of 
the  size  of  its  angles  (see  the  above  figure). 

3.  Prove  Corollary  4,  §  332,  by  moving  the  A  T  to  the  position 
of  the  A  T'  along    the    extension   of 

the  base. 

4.  Construct  a  parallelogram  with 
sides    of    2    in.  and   3  in.  and   an  in- 
cluded angle  of  30°.    Find  its  area. 

5.  Can  you  find  the  area  in  Ex.  4,  if  the  included  angle  is  45°? 

6.  Transform  a  parallelogram  into  another  parallelogram  hav- 
ing the  same  base  and  one  angle  equal  to  a  given  angle  (see  §  309). 

7.  The  lines  joining  the  mid-points  of  the  opposite  sides  of  a 
parallelogram  divide  it  into  four  equal  parallelograms. 

8.  A  rectangular  park  whose  dimensions  are  600  ft.  and  400  ft. 
is  crossed  obliquely  by  a  road  that  cuts   out  of  the   rectangle   a 
parallelogram  whose  sides  are  50  ft.  and  450  ft.    What  is  the  width 
of  the  road  ?    (See  §  238.) 

9.  A  rectangular  building  is  150  ft.  long,  100  ft.  wide,  and  120  ft. 
high.    How  many  square  feet  in  its  exterior  surface  ? 

10.  The  daily  circulation  of  a  certain  newspaper  is  75,000  copies. 
P^ach  copy  contains  20  pages,  whose  dimensions  are  18  in.  and  24 
in.    If  all  these  sheets  were  spread  out  edge  to  edge  on  a  street 
100  ft.  wide,  how  many  miles  of  the  street  would  be  covered? 


AEEAS  203 

PROPOSITIOX  II.    THEOREM 

333.    The  area  of  a  triangle  is  equal  to  half  the  product  of  its 
base  and  altitude. 


n 

Given  the  triangle  ABC,  with  the  base  b  and  the  altitude  h. 
To  prove  that  the  area  of  the  A  ABC  =  |  b  x  h. 

Proof.    1.  Complete  the  HABCD  with  AE  and  BC  as  con- 
secutive sides. 

2.  Then  the  HA  BCD  and  the  A  ABC  have  the  same  base 
and  altitude. 

3.  But  area  O  A  BCD  =  bh.  §  328 

4.  Also  A  ABC  =  A  ADC  =  1  HJABCD.  Why? 

.'.area  AABC  =  %bh. 

334.  COROLLARY.    If  T  and  T'  are  the  areas  of  two  triangles 
with  bases  b  and  b\  and  altitudes  li  and  h\  then 


(2)        =      [*  =  A'l>  (4>   r 

Give  the   statement  in   full  for  each 
equation. 

Discussion.  From  the  proposition  it  follows 
that  the  area  of  a  triangle  may  be  independent 
of  the  size  of  its  base  angles,  as  in  the  case  of 
a  parallelogram  (Ex.  2,  p.  202).  Thus,  from  the  accompanying  figure,  if 
E,  C,  and  D  are  points  on  a  line  parallel  to  AB,  then  &ABE  =  A  ABC 
,  Explain, 


204 


PLANE  GEOMETRY— BOOK  III 


EXERCISES 
PROBLEMS  OF  COMPUTATION 

1.  Find  the  area  of  a  triangle  if  the  base  b  and  the  altitude  h 
have  the  values  shown  in  the  following  table : 


I 

11 

ill 

IV 

V 

VI 

VII 

b 

h 

4 

5 

7.5 

»i 

x  +  y 

a  +  b 

V3 

7 

6 

6.5 

4£ 

x  —  y 

c  +  d 

V7 

2.  The  legs  of  a  right  triangle  are  4  and  5  (6,  7;  a  +  b,  a  —  b ;  x,  y). 
Find  the  area  in  each  case. 

3.  The  hypotenuse  of  a  right  triangle  is  10  (20,  30,  a,  x  +  y). 
Find  the  area  if  the  triangle  is  isosceles. 

4.  The  base  of  an  isosceles  triangle  is  10  v  3  ;  each  leg  is  10  ;  the 
vertex  angle  is  120°.    Find  the  area  of  the  triangle. 

5.  Each  of  the  triangular  faces  of  the  Pyramid  of  Cheops,  in 
Egypt  (p.  1),  has  a  base  of  nearly  755  ft.  and  an  altitude  of  nearly 
610ft.    How  many  square  feet  in  the  surface  of  the  pyramid? 


D 


PROBLEMS  OF  CONSTRUCTION 

6.  Transform  A  ABC  into  another  triangle  having  for  its  base 
the  line  AE,  where  E  is  any  point  on  AB  or  AB  produced. 

Solution.   Connect  C  and  E.   Draw 
BD  II  to  EC,  and  draw  DE. 

Then   &ADE  =  AABC.        Why? 

7.  Transform  a  given  triangle  into 
another  triangle  having  a  base  twice 
as  long.    What  change  takes  place  in 
the  altitude? 

8.  What  is  the  locus  of  the  vertices  of  all  equal  triangles  on  the 
same  base? 

9.  Given  two  equal  line-segments.   Required  to  construct  on  these 
segments  as  bases  equal  triangles  having  their  vertices  at  the  same 
point.    What  is  the  locus  of  this  common  vertex  ? 


A 


E 


AEEAS 


205 


10.  Transform  a  triangle  into  an  isosceles  triangle  having  the 
same  base. 

11.  Transform  a  triangle  into  another  triangle  having  two  sides 
equal  to  two  lines  c  and  d. 

12.  Construct  a  triangle  three  (four,  five,  ri)  times  as  large  as  a 
given  triangle. 

13.  Divide  a  parallelogram  into  four  (six,  three,  five)  equal  parts 
by  lines  through  one  vertex. 

14.  Given  the  triangle  ABC.   Extend  the  three  sides  in  succes- 
sion, each  by  its  own  length.   Join  the  extremities  of  the  segments 
so  constructed.    Compare  the  area  of  the  given  triangle  with  that  of 
each  triangle  in  the  figure.   What  is  the  result  if  each  side  is  ex- 
tended three  (four,  five)  times  its  own  length  ? 

15.  Transform  a  parallelogram  into  a  rhombus. 


THEOREMS 

16.  Derive  a  proof  for  Proposition  II  from  the  figure  below. 

17.  A  median  of  a  triangle  divides  it  into 
two  equal  triangles. 

18.  The  diagonals  of  a  parallelogram  divide 
it  into  four  equal  triangles. 

19.  Any  line  passing  through  the  intersec- 
tion point  of  the  diagonals  of  a  parallelogram 
divides  the  parallelogram  into  two  equal  parts. 

20.  In  the  &ABC,  D  is  any  point  in  the 
median  AE.    Prove  that  &ABD  =  &ADC. 

21.  The  point  in  which  the  three  medians 
of  a  triangle  meet  is  the  vertex  of  three 
equal  triangles  whose  bases  are  the  sides  of 
the  given  triangle. 

22.  The  area  of  an  isosceles  right  triangle  is  one  fourth  the  square 
on  the  hypotenuse. 

23.  If  the  mid-points  of  two  sides  of  a  triangle  are  joined,  a  triangle 
is  formed  which  is  one  fourth  the  given  triangle. 

24.  The  area  of  a  circumscribed  polygon  is  equal  to  half  the  product 
of  its  perimeter  and  the  radius  of  the  inscribed  circle. 


206 


PLANE  GEOMETRY— BOOK  III 


PROPOSITION  III.    THEOREM 

335.    The  area  of  a  trapezoid  is  equal  to  half  the  product 
of  its  altitude  and  the  sum  of  its  bases. 

I' 


Given  the  trapezoid  ABCD,  with  the  bases  b  and  6'  and  the 
altitude  h. 

To  prove  that  the  area  of  A  BCD  =  1  h  (b  +  b'). 

1.  Draw  the  diagonal  A  C. 

2.  Then  A  ABC,  A  A  DC,  and  trapezoid  A  B  CD  have  the  same 
altitude. 

3.  But  area  A  ABC  =  J  bh,  §  333 

and  -  area  A  A  DC  =  £  b'h. 

4.  /.  area  trapezoid  A  BCD  =  %k(b  +  bt).  Ax.  2 

336.   COROLLARY.    The  area  of  a  trapezoid  is  equal  to  the 
product  of  its  altitude  and  mid-line.  §  219 


EXERCISES 

1.  From  §  321  derive  a  proof  for  the  above  theorem  by  means  of 
tha  figure  below.  .  r 

2.  In  the  laying  of  a  railway  track 
an  excavation  had  to  be  made.  A  ver- 
tical cross   section  of  the  excavation 
perpendicular  to  the  roadbed  had  the 
form  of  a  trapezoid  with  bases  of  50  ft. 
and  68  ft.  The  depth  of  the  excavation 

was  12  ft.  and  its  length  was  400  ft.    Find  the  area  of  the  cross 
section,   How  many  cubic  yards  of  dirt  had  to  be  removed  ? 


AREAS 


207 


3.  The  dimensions  of  a  picture  are  15  in.  and  10  in.    The  picture 
is  surrounded  by  a  mat  4  in.  wide.    What  is  the  area  of  the  mat  ? 

4.  The  diagram  shows  how  the  area  of 
an  irregular  polygon  may  be  found,  if  the 
distance  of  each  vertex  from  a  given  base 
line,  as  XY,  is  known.    These  distances 
A  A',  BB',  etc.,  are  called  offsets,  and  are 
the  bases  of  trapezoids  whose  altitudes  are 
A'B',  B'C',  C'D',  etc.  The  area  ABCDEF 
may  now  be  found  by  the  proper  additions 
and  subtractions. 

On  cross-section  paper  plot  the  points 
whose  coordinates  are  given  below,  join 
them  in  order,  draw  FA,  and  find  the  inclosed  area  in  each  case: 


A 

R 

G 

l) 

£ 

F 

3,  6 

2,4 

3,0 

5,  1 

3,  2 

4,  5 

3,  7 

1,  3 

4,0 

6,1 

6,  6 

5,7 

2,6 

3,4 

1,  2 

4,1 

5,5 

3,7 

5.  In  order  to  determine  the  flow  of  water  in  a  certain  stream, 


Add 


d     d     d  B 


soundings  are  taken  every  6  ft.  on  a  line 
AB  at  right  angles  to  the  current.  A 
diagram  may  then  be  made  to  repre- 
sent a  vertical  cross  section  of  the 
stream.  If  the  area  of  this  cross  sec- 
tion and  the  speed  of  the  current  are 
known,  it  is  possible  to  determine  the  amount  of  water  flowing 
through  the  cross  section  in  a  given  time. 

The  required  area  is  often  found  approximately  by  joining  the 
extremities  of  the  offsets  y0,  yv  yv  etc.,  by  straight  lines,  and  finding 
the  sum  of  the  trapezoids  thus  formed.  That  is,  the  strips  between 
successive  offsets  are  replaced  by  trapezoids.  This  gives  the  Trape- 
zoidal Rule  for  finding  an  area.  It  may  be  stated  as  follows  :  To  half 
the  sum  of  the  first  and  last  offsets  add  the  sum  of  all  intermediate  offsets, 
and  multiply  this  result  by  the  common  distance  between  the  offsets. 

Prove  this  rule. 


208 


PLANE  GEOMETKY— BOOK  III 


6.  Find  the  area  of  the  cross  section  of  a  stream  if  the  soundings 
taken  at  intervals  of  6  ft.  are  respectively  4  ft.,  5|  ft.,  10  ft.,  12^  ft., 
15|  ft.,  8£  ft.,  and  6  ft. 

7.  Show  how  the  Trapezoidal  Rule  may  be 
applied  to  find  approximately  the  area  of  a 
figure  bounded  by  a  curved  line. 

8.  In  the  midship  section  of  a  vessel  the  width  taken  at  intervals 
of  1  ft.  is  successively  17,  17.2,  17.4,  17.4,  17.4,  17.2,  16.8,  16,  14,  8, 
and  2,  measurements  being  in  feet.    Find  the  area  of  the  section. 
(Use  the  line  drawn  from  the  keel  _1_  to  the  deck  as  base  line.) 

9.  A  second  rule  for  finding  plane  areas,  known  as  Simpson's  Rule, 
usually  gives  a  closer  result  than  the  Trapezoidal  Rule.    In  proving 
Simpson's   Rule  two  consecutive  strips   are 

replaced  by  a  rectangle  and  two  trapezoids  as 
follows :  Divide  2  d  into  three  equal  parts, 
erect  Js  at  the  points  of  division,  and  complete 
the  rectangle  whose  altitude  is  the  middle 
offset  yv  as  in  the  figure.  Join  the  extrem- 
ities of  y  and  y2  to  the  nearer  upper  vertex 
of  this  rectangle.  Then  if  the  areas  of  the  trapezoids  are  T  and  T', 
and  if  the  area  of  the  rectangle  is  R, 


If,  now,  the  number  of  strips  is  even,  and  if  the  offsets  are  lettered 
consecutively  y0,  yv  y2,  .  .  .,  yn,  the  addition  of  the  areas  of  successive 
double  strips,  found  by  the  above  formula,  gives  the  result 


Area  = 


+  4  ^  +  2  y2  +  4y8  +  •  •  •  +  2yn_ 


_1  +  yn}. 


In  words  :  To  the  sum  of  the  first  and  last  offsets  add  twice  the  sum 
of  all  the  other  even  offsets  and  four  times  the  sum  of  all  the  odd  offsets, 
and  multiply  by  one  third  the  common  distance  between  the  offsets. 

Solve  Exs.  6  and  8  by  Simpson's  Rule. 

10.  Find  the  area  of  a  piece  of  steel  plate  which  has  an  axis  of 
symmetry,  if  the  offsets  measured  on  each  side  from  this  axis  at 
intervals  of  20  cm.  are  successively  54,  68,  72,  72,  60,  44,  36,  28,  and 
20,  measurements  being  in  centimeters. 


AREAS 


209 


PROPOSITION  IV.    THEOREM 

337.  If  two  triangles  have  an  angle  of 'one  equal  to  an  angle 
of  the  other,  their  areas  are  to  each  other  as  the  products  of  the 
sides  including  the  equal  angles. 


C'  E 

Given  two  triangles  ABC  and  A'B'C',  having  the  angle  A  equal 
to  the  angle  A'. 

To  prove  that 


AABC 


AA'B'C' 


bc_ 
b'c1' 


Proof.  1.  Produce  B  'A '  through  A '  to  E,  so  that  A  'E  =  A  C,  and 
produce  C  'A '  through  A '  to  D,  so  that  A'D= AB.  Draw  ZXE  and  £'Z). 


2.  Then 

3.  ~Now 

and 
4. 

that  is, 


AA'DE  =  AABC. 
AA'DE   b 


AA'DB1 
AA'DB' 
AA'B'C' 

AA'DE 
AA'B'C' 

AABC 
AA'B'C' 


c 

G 
V 

be 

W 

bc_ 
b'c'' 


s.  a.  s. 
§  334,  (2) 

Why? 
Ax.  4 
Ax.  1 


338.  COROLLARY.  The  bisector  of  an  interior  angle  of  a  tri- 
angle divides  the  opposite  side  into  segments  which  are  to  each 
other  as  the  adjacent  side's  of  the  triangle. 

AT      at      a 


AT' 
AT 


Suggestion. 
But  also 


This  corollary  enables  us  to  find  the  segments 
determined  by  the  bisector  t  upon  the  opposite  side. 


210  PLANE  GEOMETEY— BOOK  III 

TRANSFORMATIONS 

PROPOSITION  V.    PROBLEM 
339.  To  transform  a  quadrilateral  into  a  triangle. 


D 

Given  the  quadrilateral  ABCD. 
Required  to  construct  a  triangle  equal  to  ABCD. 

Construction.  Draw  the  diagonal  BD,  and  draw  CE  II  to  BD, 
meeting  the  side  AD  produced  in  E.  Draw  BE.  Then  BE  A  is 
the  required  triangle. 

Proof.    1.  ABCD  =  ABDE.  Why? 

2.  .  • .  quadrilateral  ABCD  =  A  BE  A .  Why  ? 

340.  COROLLARY.  To  transform  a  pentagon  into  a  triangle, 
first  transform  it  into  a  quadrilateral  by  the  above  method, 
and  then  transform  the  quadrilateral  into  a  triangle.  A  poly- 
gon of  any  number  of  sides  may  be  transformed  into  a  triangle 
by  a  repetition  of  this  process. 

EXERCISES 

1.  In  how  many  ways  may  a  quadrilateral  be  transformed  into  a 
triangle  ? 

2.  Draw  on  a  large  scale  an  irregular  hexagon,  and  transform  it 
into  a  triangle.   Measure  the  base  and  the  altitude  of  this  triangle, 
and  compute  its  area.    Why  is  Proposition  V  of  importance  ? 

3.  In  how  many  different  ways  can  a  pentagon  be  transformed 
into  a  triangle  by  the  above  method  ? 

4.  Transform  a  square  into  a  right  triangle;  into  an  isosceles 
triangle. 


TRANSFORMATIONS  211 

PROPOSITION  VI.    THEOREM 

341.  If  through  a  point  on  a  diagonal  of  a  parallelogram 
parallels  to  the  sides  are  drawn,  the  parallelograms  formed  on 
the  opposite  sides  of  that  diagonal  are  equal. 


Given  the  parallelogram  ABCD,  and  through  P,  a  point  on  AC, 
the  line  EF  parallel  to  AB,  and  GH  parallel  to  AD,  forming  the 
parallelograms  b  and  &',  and  the  triangles  a,  ar,  c,  and  d. 

To  prove  that  H  b  =  O  b'. 

Proof.    1.  A  (a  +  I  +  c)  =  A  (a1  +  V  +  c').  Why  ? 

2.  But  A  a  =  A  a'  and  A  c  =  A  c'.  Why  ? 

3.  .'.nb  =  nV.  Why? 
This  theorem  aids  in  proving  some  important  constructions. 

342.  COROLLARY  1.  To  transform  a  parallelogram  into  an- 
other parallelogram  whose  angles  are  equal  to  those  of  the  given 
parallelogram  and  ivhose  base  is  equal  to  a  given  line. 


p  ' 


Lay  off  the  new  base  upon  the  extension  of  the  original,  and  complete 
the  figure  as  shown.    Then  P*=  P'. 

343.   COROLLARY  2.   To  transform  a  triangle  into  another  tri- 
angle having  a  given  base.   (Second  Method.    See  p.  204,  Ex.  6.) 

Since  any  triangle  may  be  regarded  as  one  half  of  a  parallelogram,  the 
above  construction  may  be  extended  to  include  triangles. 


212 


PLANE  GEOMETRY— BOOK  III 


EXERCISES 
SIMPLE  TRANSFORMATIONS 

1.  Transform  a  A  AB C  into  a  triangle  with  its  base  on  BC,  or  EC 
extended,  and  with  the  opposite  vertex  (a)  on  AC]  (b)  on  AB  ex- 
tended; (c)  within  &ABC;  (d)  without  &ABC.   (See  Ex.  6,  p.  204.) 

2.  Transform  a  rectangle  into  (a)  another  rectangle  having  a 
different  base ;  (b)  a  parallelogram  of  given  base  and  given  angles ; 
(c)  a  parallelogram  of  given  sides ;  (d)  a  triangle  of  given  base ; 
(e)  a  trapezoid  of  the  same  altitude  and  a  given  base. 

3.  Transform  a  parallelogram  into  (a)  a  rectangle  of  given  alti- 
tude ;  (b)  a  parallelogram  of  given  altitude ;  (c)  a  triangle  of  given 
altitude ;  (d)  a  trapezoid  of  the  same  altitude. 

4.  The  figures  show  easy  methods  of  transforming  (a)  a  triangle 
into  a  parallelogram  ;   (b)  a  parallelogram  into  a  triangle ;   (c)  a 
trapezoid  into  a  parallelogram.    Explain. 


C   E 


F  D 


DIVISIONS  AND  DISSECTIONS 

5.  From  one  vertex  of  a  triangle  draw  lines  dividing  the  triangle 
into  (a)  three  (five,  six)  equal  parts ;  (b)  two  parts  that  shall  be  in 
the  ratio  of  3  : 1  (4  : 1 ;  3  : 4)  ;  (c)  three  parts  that  shall  be  in  the  ratio 
of  1:2:3  (2:3:4). 

6.  Solve  Ex.  5  if  a  parallelogram  is  given  instead  of  a  triangle. 

7.  Divide  a  trapezoid  into  two  (three,  five)  equal  trapezoids. 
Suggestion.  Divide  each  of  the  bases  into  as  many 

equal  parts  as  the  trapezoid  is  to  have. 

8.  A  polygon  is  divided  into  two  parts,  P  and 
P',  by  the  broken  line  ABC.    Explain  how  ABC 
may  be  replaced  by  a  straight  line  so  as  not  to  in- 
crease the  area  of  either  P  or  P'. 


TRANSFORMATIONS 


213 


9.  Divide  a  quadrilateral  into  three  equal  parts  by  drawing  lines 
from  one  vertex. 

Suggestion.  Transform  the  quadrilateral  into  a  triangle  having  that 
vertex  as  its  vertex,  and  proceed  as  in  Ex.  5. 

10.  Show  at  least  six  ways  of  dividing  a 
given  parallelogram  into  four  equal  parts. 

11.  In  the  accompanying  rectangular  fig- 
ure,   representing    an    ornamental    window, 


what  part  of  the  total  area  is   Aa? 
Oc?   &ABC1   AADE't   OMNOP'l 


Ai? 


12.  Transform  into  a  triangle,  and  then  into  a  rectangle,  (a)  a 
pentagon ;  (b)  an  octagon. 


ADDITION  OF  TRIANGLES  AND  PARALLELOGRAMS 

13.  Construct  a  triangle  equal  to  the  sum  of  two  given  triangles. 
Suggestion.  Transform  the  first  of  the  given  triangles  into  another 

triangle  having  its  base  equal  to  the  base  of  the  other  given  triangle. 

14.  Given  the  triangle  ABC.    If  a  line-  ™ 
segment  equal  to  AD  moves  along  AB  and 

BC,  remaining  constantly  parallel  to  AD, 
prove  that  O  AE  +  a  EC  =  O  AF.   (Prove 

A  A  BC  =  A  DEF  by  s.  s.  s.) 

15.  Show  that  the  principle  of  Ex.14  ap- 
plies to  a  broken  line  ABCDE. 

16.  Theorem    due    to    Pappus 

(300  A.D.).    If  on  two  sides  of  a 
triangle  ABC  any  desired  paral- 
lelograms AE  and  BE'  are  con- 
structed, and  the  sides  DE  and 
D'E'  are  produced  to  meet  at  M,  and  on 
AC  a  parallelogram  is  constructed  having 
a  side  AG  equal  and  parallel  to  BM,  then 
OAH  =  aAE  +  aBE'.    (Cf.  Ex.  14.) 

17.  Construct  a  parallelogram   equal   to 
the  sum  of  two  given  parallelograms. 


214 


PLANE  GEOMETRY— BOOK  III 


THEOREM  OF  PYTHAGORAS 
PROPOSITION  VII.    THEOREM 

344.  In  a  right  triangle  the  square  on  the  hypotenuse  is 
>qual  to  the  sum  of  the  squares  on  the  legs. 


'£2 

X 

c  / 

^p 

1 

/  c2 

' 

/ 

L   E 

Given  the  right  triangle  ABC,  having  the  legs  a  and  b 
hypotenuse  c,  with  the  square  AE  on  the  hypotenuse  AB1 
squares  CF  and  CK  on  the  legs. 

To  prove  that  c2  =  a2  +  b2. 

Proof.  1.    Draw  CL  II  to  AD.    Draw  BK  and  CD. 
2.  Now        ACG  and  ECU  are  straight  lines. 
AACD==AAKB. 


and  the 
and  the 


3.  Also 
For 

and 

4.  But 
and 

5. 


=  AD, 


area  rect.  AL  =  2  area  A.4CZ>, 
area  square  CK  =  2  area  A  A  KB. 
.'.  rect.  ^4i  =  square  C/v. 


Why? 
s.  a.  s. 
Why? 

Why? 
Why? 

Ax.  1 


THEOREM  OF  PYTHAGORAS 


215 


6.  In  like  manner,    by  drawing  CE  and  AF,  it  may  be 
proved  that 


rect.  BL  =  square  CF. 


1.  But 


Ax.  7 
Ax.  1 

c*=a*  +  b\  Why? 

The  square  on  one  leg  of  a  right  triangle 


square  AE  =  rect.  AL  -f-  rect.  BL. 
.'.  square  AE  =  square  CK  -f-  square  CF. 
That  is,  c2  =  a2  +  £2. 

345.  COROLLARY. 


is  equal  to  the  square  on  the  hypotenuse  diminished  by  the  square 
on  the  other  leg. 

346.  Historical  Note.    Proposition  VII  is  called  the  Pythago- 
rean theorem,  as  its  discovery  is  usually  credited  to  Pythagoras 
(about   550   B.C.).     It   is    practically   certain, 

however,  that  the  theorem  was  known  to  the 
Egyptians  and  the  Hindus  long  before  that 
time. 

The  Pythagorean  proposition  is  in  many 
respects  the  most  famous  single  truth  of  plane 
geometry.  It  is  of  the  greatest  importance,  and 
many  different  proofs  for  it  have  been  given. 
The  one  reproduced  above  is  due  to  Euclid 
(300  B.C.). 

347.  Alternative  Proofs.    Fig.  1  shows  how  the 
proof  might  have  been  suggested  in  the  laying  of  tile 
floors. 

Fig.  2   is  thought  by  some  writers  to  have 
been  used  by  Pythagoras.    If  A  a,  6,  c,  d,  are 
taken  from  the  large  square,  the  square  on  the 
hypotenuse  remains,  and  if  the  rectangles  AC         ^' 
and  CB  are  removed,  the  squares  on  the  legs      ^ 
remain.    But  the  two  rectangles  are  together       \ 
equal  to  the  four  triangles.    Hence  the  remain-         \ 
ders  are  equal.  *T\ 


FIG.  2 


FIG.  3 


348.   Fig.  3  is  used  in  a  proof  ascribed  to 
President  Garfield.    On  the  hypotenuse   con- 
struct one  half  of  a  square,  and  draw  of  per- 
pendicular to  a  produced.    Prove  that  A  abc  =  A  a'b'c',  and  that  the 
entire  figure  is  a  trapezoid  of  altitude  a  +  &',  etc. 


216  PLANE  GEOMETRY— BOOK  III 


EXERCISES 
NUMERICAL  PROBLEMS 

(A  knowledge  of  how  to  extract  the  square  root  of  a  number  is  neces- 
sary for  the  following  problems.  Results  should  be  obtained  correct  to 
two  decimal  places,  unless  otherwise  specified.  If  it  is  required  to  cal- 
culate the  length  of  one  leg  a  of  a  right  triangle  when  the  other  leg 
6  and  the  hypotenuse  c  are  given,  it  will  be  found  helpful  to  bear  in 
mind  that  the  equation  a2  =  c2  —  &2  may  be  written  a2  =  (c  +  6)  (c  —  6). 
For  example,  if  c  =  37,  6  =  35,  then  a2  =  372  -  352  =  (37  +  35)  (37  -  35) 
=  72x2  =  144.  .-.  a  =12.) 

1.  If  a?  and  y  in  the  following  table  represent  the  lengths  of  the 
legs  of  a  right  triangle,  find  the  length  of  the  hypotenuse  in  each  case  : 


X 

4 

12 

24 

a 

a  +  b 

y 

3 

5 

7 

b 

c  +  d 

2.  If,  in  Ex.  1,  x  represents  the  hypotenuse  and  y  one  of  the  legs, 
find  the  other  leg. 

3.  Find  the  diagonal  of  a  rectangle  whose  dimensions  are  8  and 
15  ;  11  and  60  ;  13  and  84. 

4.  A  parallelogram  is  inscribed  in  a  circle  whose  diameter  is  13. 
One  side  of  the  parallelogram  is  5.   Find  the  other  side. 

5.  The  base  of  an  isosceles  triangle  is  4,  and  its  legs  are  each  5. 
Find  the  altitude  and  the  area. 

6.  The  side  of  an  equilateral  triangle  is  6.    Find  the  area. 

7.  The  bases  of  an  isosceles  trapezoid  are  6  and  12,  and  the  legs 
are  each  5.    Find  the  area. 

8.  The  diagonals  of  a  rhombus  are  6  and  8.  Find  its  perimeter. 

9.  A  chord  48  in.  long  is  7  in.  from  the  center  of  its  circle.    How 
long  is  the  radius  ? 

10.  If  the  coordinates  of  two  points,  P  and  Q,  are  (2,  1)  and 
(6,  4)  respectively,  find  the  length  of  the  line  PQ. 

11.  If  the  coordinates  of  the  vertices  of  a  triangle  are  (1,  1), 
(3,  3),  and  (1,  5),  find  the  perimeter  of  the  triangle ;  find  its  area. 


THEOREM  OF  PYTHAGORAS  217 

12.  Pythagorean  Numbers.    Prove  that  the  three  integers  in  each 
of  the  following  groups  may  represent  the  sides  of  a  right  triangle. 

(a)  3,  4,  5.  (d)  8,  15,  17. 

(b)  5,  12,  13.  (e)  11,  60,  61. 

(c)  7,24,25.  (f)  12,35,37. 

NOTE.    The  above  list  may  be  extended  by  the  use  of  the  following 
formulas  : 

(1)  If  the  length  of  a  leg  is  denoted  by  an  even  integer  n,  the  lengths 

nz  _  4         n2  +  4 
of  the  other  two  sides  are  —     —and—     —    (Plato.) 

4       n2  —  1          n2  +  1 


(2)  If  n  is  odd,  the  other  two  integers  are  --  and 


(Pythagoras.) 

13.  The  centers  of  two  circles  are  25  in.  apart.  Their  radii  are  3  in. 
and  10  in.  respectively.    Find  the  length  of  their  common  external 
tangent  (see  p.  161). 

14.  The  centers  of  two  circles  are  13  in.  apart.    If  their  radii  are 
3  in.  and  2  in.  respectively,  what  is  the  length  of  the  common  internal 
tangents?    (See  p.  161.) 

APPLIED  PROBLEMS 

1.  What  is  the  diagonal  of  a  rectangular  floor  whose  dimensions 
are  12  ft.  and  9  ft.  ? 

2.  Find  how  far  a  pedestrian  is  from  his  starting  point  if  he 
walks  (a)  12  mi.  N.  and  then  5  mi.  E.  ;   (b)  20  mi.  S.E.  and  then 
21  mi.  S.W.  ;   (c)  5  mi.  N.,  3  mi.  E.,  2  mi.  N.  ;   (d)  2  mi.  W.  and  then 
10  mi.  S.W. 

3.  A  captive  balloon  rises  vertically  to  a  height  of  1000  ft.    How 
far  is  it  from  an  observer  400  ft.  away  from  the  point  of  ascent  ? 

4.  A  ladder  25  ft.  long  reaches  to  a  window  24  ft.  from  the 
ground.    How  far  is  the  foot  of  the  ladder  from  the  wall? 

5.  Two  telephone  poles  are  60  ft.  apart.    The  height  of  the  poles 
is  40  ft.  and  30  ft.    How  long  is  a  wire  connecting  their  tops  ? 

6.  A  ladder  50  ft.  long  is  placed  against  a  window  40  ft.  from  the 
ground.    The  upper  extremity  of  the  ladder  slides  down  1  ft.    How 
far  does  the  other  extremity  of  the  ladder  slip  ? 


218 


PLANE  GEOMETRY— BOOK  III 


12 


10 


7.  A  derrick  has  a  movable  arm  41  ft.  long.    A  weight  to  be 
lifted  is  9  ft.  from  the  foot  of  the  arm.    How  long  is  the  cable  ex- 
tending from  the  end  of  the  arm  to  the  weight? 

8.  The  side  of  a  square  baseball  field  is  90  ft.    Find  the  distance 
from  second  base  to  the  home  plate. 

9.  A  hexagonal  floor  has  the  form  and  the 
dimensions  indicated  in  the  figure.   Find  the  area. 

10.  An  elevated  train  moving  at  the  rate  of  25  mi. 
an  hour  passes  above  a  surface  car  going  at  the  rate 
of  15  mi.  an  hour.    The  two  tracks  cross  at  right 

angles.  Find  the  distance  separating  the  train  and  the  car  after  10  min. 

11.  A  coast-defense  gun  has  a  range  of  10  mi.    A  boat  sails  along 
the  shore  at  a  distance  of  8  mi.,  going  at  the  rate  of  18  mi.  per  hour. 
How  long  is  the  boat  within  the  range  of  the  gun  ? 

12.  A  pendulum  is  39.1  in.  long.    Its  bob  is  drawn  to  one  side  un- 
til it  is  1  in.  higher  than  it  was  originally.    How  far  is  it  from  the 
vertical  line  passing  through  the  point  of  support  of  the  pendulum? 

13.  A  teeter  board  12  ft.  long  is  supported  at  its  center  by  a  frame 
3  ft.  high.    How  high  can  each  extremity  of  the  board  rise  ? 

14.  A  box  has  the  form  of  a  cubic  meter.    How  many  millimeters 
in  the  diagonal  of  the  box  ? 

15.  The  dimensions  of  a  rectangular  room  are  16  ft.,  12  ft.,  and 
9  ft.    How  long  is  a  line  connecting  a  lower  corner  with  the  opposite 
upper  corner  ? 

16.  The  diagram  represents  a  lever 
OA,  2  ft.  long,  which  may  revolve  about     m 
O  in  a  vertical  plane.    In  the  plane 

of  its  revolution  it  is  made  to  push 
against  a  horizontal  rod  EC  attached 
to  a  vertical  rod  DE.  B  projects  4  in. 
beyond  A.  DE  is  compelled  to  move 
vertically  by  the  guides  m  and  n.  As  OA 
revolves,  how  high  is  DE  raised  by  it  ? 

17.  A  tree  is  broken  24  ft.  from  the  ground.    The  two  parts  hold 
together  and  the  top  of  the  tree  touches  the  ground  7  ft.  from  the 
foot.    Find  the  height  of  the  tree. 


D 


THEOREM  OF  PYTHAGOKAS 


219 


18.  In  the  middle  of  a  pond  10  ft.  square  grew  a  reed.    The  reed 
projected  1  ft.  above  the  surface  of  the  water.    When  blown  aside  by 
the  wind,  its  top  part  reached  to  the  mid-point  of  a  side  of  the  pond. 
How  deep  was  the  pond?    (Old  Chinese  problem.) 

19.  The  width  of  a  house  is  28  ft.    The  triangle  in  the  gable  is 
right-angled.     How  long  is  each   of    the   rafters  in  this  triangle  ? 
What  is  the  area  of  the  triangle  ?    How  high  is  it  ? 

20.  A  house  is  32  ft.  wide,  45  ft.  long,  25  ft.  high  to  the  roof  and 
35  ft.  high  to  the  ridgepole.    How  many  square  feet  are  there  in  its 
entire  exterior  surface? 

21.  A  circular  pavilion  has  a  conical  roof.    The  pavilion  is  20  ft. 
high  to  the  roof,  and  30  ft.  high  to  the  vertex  of  the  cone.    The  di- 
ameter of  the  pavilion  is  25  ft.    The  flagpole  is  12  ft.  high.    How 
long  is  the  guy  rope  joining  the  top  of  the  flagpole  to  the  edge  of 
the  roof  ? 

22.  Masons   and  carpenters,  in   laying  out   the    plan    of    a    rec- 
tangular   building,     often     use    the    following 

method  of  constructing  the  right  angles  at  the 
corners  :  Three  poles,  whose  lengths  are  6  ft.,  8  ft., 
and  10  ft.  respectively,  are  joined  together  at 
their  extremities  so  as  to  form  a  triangle.  The 
angle  opposite  the  largest  pole  is  a  right  angle. 
(Why  ?  )  A  similar  method  is  known  to  have 
been  used  by  the  surveyors,  or  "  rope-stretchers,"  of  ancient  Egypt, 

23.  The    theorem    of    Pythagoras 
is   employed    to   find    the   "equation 
of   a   circle "  about  the   origin  as  a 
center. 

Take  any  point  P  in  a  circle  about 
the  origin  O.  Draw  the  ordinate  HfP. 
Let  OM  =  x,  and  MP  =  y.  Then 
~OMi  +  MP'2  =  OP2.  If  the  radius 
OP  =  r,  this  becomes  x2  +  y'2  =  r2. 
This  equation  holds  for  the  coordi- 
nates of  any  point  on  the  circle,  and  is  called  the  equation  of  the 
circle,  r  being  any  known  number. 

24.  Plot  the  equations  x2  +  f  =  25  ;  x2  +  f  =  98.  (See  Ex.  23.) 


10 


,<s 

A 

B 

a       V 
\ 

\ 

a 

b 

\ 

220  PLANE  GEOMETRY— BOOK  III 

PROPOSITION  VIII.    PROBLEM 

349.   To  construct  a  square  equal  to  the  sum  or  the  difference 
of  two  given  squares. 


\a 


-c-'-l 

c'  ' 

u ! 


Given  two  squares  A  and  B. 

Required  to  construct  a  square  C  equal  to  A  -\-  B,  and  a  square 

C1  equal  to  A  —  B. 

(To  be  completed.) 

EXERCISES 

1.  Construct  a  square  equal  to  the  sum  of  three  given  squares. 

2.  If  a,  &,  and  c  are  line  segments,  construct 

x  =  Va2  +  b2 ;  x  =  V«2  +  b2  +  c2. 

3.  Construct  a  square  twice  as  large  as  a  given  square ;   three 
times  as  large. 

4.  If    squares  be  constructed  on  the  diagonals  of   a  rectangle, 
prove  that  their  sum  equals  the  sum  of  the  squares  constructed  on 
the  sides  of  the  rectangle. 

5.  Represent  by  squares  the  areas  of  the  following  countries  and 
states,  using  a  scale  in  which  1  cm.  represents  100  mi. : 

United  States  (including  Alaska)     .     .     .  3,605,600  sq.  mi. 

Texas 265,780  sq.  mi. 

California 158,360  sq.  mi. 

New  York 49,170  sq.  mi. 

6.  Represent  by  three  squares  the  area  of  the  state  in  which  you 
live  and  of  two  neighboring  states.    Then  construct  a  square  repre- 
senting the  combined  areas  of  the  three  states. 


THEOREM  OF  PYTHAGORAS  221 

PROPOSITION  IX.    PROBLEM 

350.   To  construct  a  square  equal  to  a  given  rectangle  (or 
parallelogram).  G 

/     \  v 


JOl 

E 


D       G 

Given  the  rectangle  ABCD. 

Required  to  construct  a  square  equal  to  ABCD. 

Construction.  1.  Produce  AB,  the  shorter  side  of  ABCD, 
to  E,  making  AE  =  AD,  and  on  AE  as  a  diameter  construct 
a  semicircle. 

2.  Construct  BF  _L  to  AE,  meeting  the  semicircle  at  F. 

3.  Draw  AF.    The  square  on  AF  is  the  required  square. 
Proof.  1.  Draw  HE  and  DF. 

2.  Then  AADF=AAHE.  s.  a.  s. 

3.  But  area  rect.  AC  =  2  area  A  A  DF,  Why  ? 
and                   area  square  FH  =  2  area  A  A  HE. 

4.  Hence  rect.  ABC D=  square  HF.  Ax.  1 

351.  COROLLARY  1.   To  transform  a  triangle  into  a  square, 
first  transform  the  triangle  into  a  rectangle  having  the  same  base 
as  the  triangle  and  an  altitude  equal  to  one  half  the  altitude  of 
the  triangle,  and  then  transform  that  rectangle  into  a  square. 

352.  COROLLARY  2.    To  transform  a  polygon  into  a  square, 
first  transform  the  polygon  into  a  triangle  (§  340),  and  then 
transform  that  triangle  into  a  square,  as  above. 

NOTE.  Proposition  IX  enables  us  to  find  the  area  of  any  polygon  by 
one  measurement,  namely,  that  of  the  side  of  the  equal  square. 


222  PLANE  GEOMETRY— BOOK  III 

PROPOSITION  X.    PROBLEM 

353.    To  construct  a  square  which  shall  be  a  given  part  of 
a  given  square.  H 


/ 

\\ 

D      E  C 

Given  the  square  ABCD. 

Required  to  construct  a  square  equal  to  %  of  ABCD. 
Construction.    1.  On  AB  construct  BF=%  AB.  §  212 

2.  Draw  EF\\  to  AD. 

3.  Transform  the  rectangle  EFBC  into  a  square  BH  (§  350). 
This  is  the  required  square. 

(Proof  to  be  completed.) 


EXERCISES 

CONSTRUCTIONS 

1.  Construct  a  square  equal  to  a  triangle  of  base  7cm.  and  alti- 
tude 5  cm. 

2.  Construct  \/21. 

Solution.  Let  V21  =  x.  Then  x2  =  21  =  7-3.  (Apply  §  350.) 

3.  Construct    Vl2  in.  by  several  methods.    Compare  your  final 
results. 

4.  Construct  x2  —  %  a2.    (Suggestion*   x2  =  \  a  •  «.) 

5.  Transform  a  pentagon  into  a  square. 

6.  Transform  an  isosceles  triangle  into  a  square. 

7.  Construct  a  square  equal  to  £  of  a  given  hexagon. 

8.  Construct  a  square  equal  to  the  sum  of  two  given  triangles. 


THEOREM  OF  PYTHAGORAS  223 

NUMERICAL  PROBLEMS 

1.  If  the  side  of  a  square  is  a  and  its  diagonal  is  d,  show  that 

d  —  a  V2  and  that  a  =  — -  • 

V2 

2.  The  diagonal  of  a  square  is  28  cm.    Find  the  length  of  a  side. 

28       28      28      280 
Solution,    a  =  —=  =  —-  =  —  =  ——  =  20,  approximately. 

3.  If  the  hypotenuse  of  an  isosceles  right  triangle  is  c,  show  that 

c2 
the  area  of  the  triangle  is  —  • 

4.  The  length  of  the  leg  of  an  isosceles  right  triangle  is  7.    What 
is  the  area  ?   What  is  the  hypotenuse  ? 

5.  The  hypotenuse  of  an  isosceles  right  triangle  is  10.    What  is 
the  area  ?    How  long  is  each  leg  ? 

«2    /- 

6.  If  a  side  of  an  equilateral  A  is  «,  show  that  the  area  is  —  V3. 

7.  Find  the  area  of  an  equilateral  A  in  terms  of  its  altitude. 

8.  The  side  of  an  equilateral  triangle  is  6.   Find  the  altitude  and 
the  area. 

9.  A  side  of  a  regular  hexagon  is  8.    What  is  the  area  of  the 
hexagon  ? 

10.  A  regular  hexagonal  prism  10  in.  high  has  base  edges  of  4  in. 
How  many  square  inches  in  the  entire  exterior  surface  ? 

11.  The  base  edges  of  a  pyramid  with  a  square  base  are  5  in.  in 
length,  the  height  of  the  pyramid  being  7  in.    Find  the  length  of 
the  other  edges. 

12.  A  triangular  pyramid  is  formed  by  4  congruent  equilateral 
triangles  of  side  10  cm.    Such  a  pyramid  is  called  a  regular  tetrahe- 
dron.   How  many  square  centimeters  in  its  entire  surface  ? 

13.  A  right  triangle  has  legs  of  6  cm.  and  8  cm.   Find  the  altitude 
on  the  hypotenuse. 

14.  A   rhombus  has    diagonals   of    12  in.   and    18  in.     Find   its 
altitude. 

15.  A  rhombus  whose  side  is  5  in.  has  acute  angles  of  60°.    What 
is  its  area  ? 


224 


PLANE  GEOMETRY— BOOK  III 


16.  What  is  the  result  in  Ex.  15,  if  the  acute  angles  are  45°? 

17.  A  square  frame  consists  of  four  sticks  hinged  together  at 
their  extremities.    The  length  of  each  stick  is  8  in.    The  frame  is 
drawn  to  one  side  until  two  of  the  angles  become  120°.    Is  the  area 
increased  or  decreased,  and  how  much  ? 

18.  A  square  frame,  20  in.  on  a  side,  is  to  be  stiffened  by  a  diag- 
onal crosspiece.    How  long  is  the  crosspiece  ? 

19.  The  side  of  a  square  is  5  cm.    Equal  dis- 
tances of  1  cm.  are  laid  off  on  the  sides  from  the 
vertices.    Find  the  length  of  RS;  of  ST-,  of  MN. 
Find  the  area  of  RSTV;  of  MNOP;  of  RMW. 

20.  A  staircase  has  11  stairs,  each  2  a  in.  wide 

and  a  in.  high.    How  long  a  board  will  completely  protect  it  ? 

21.  The  edge  of  a  cube  is  1  in.    Find  its  diagonal. 

22.  The  dimensions  of  a  rectangular  room  are  a,  b,  and  c.    How 
long    is    a    diagonal    from   a    lower  corner  to    the    opposite    upper 
corner  ? 

SQUARE  ROOTS 

23.  If  AB  =  1,  and  BC  =  1,  AC  is  a  geometric  representation  of 
V2.    Prove. 

24.  If   DE  =  1,  and  EF  =  V2,     A 
prove  that  DF  =  V3. 

25.  Construct  VS,  V5,  V?,  using 
squared  paper. 

26.  The  following  figures  illustrate  two  short  methods  of  repre- 
senting square  roots  of  small  integers.    Explain. 

B        i       C     D   E  F  AID  E       F 


=  BC=l. 

=  BD=  V2. 

=  BE=  V3,  etc. 


B  C     G        H 

A  C  is  a  square. 
BD  =  BG=  V2. 
AG  =  AE=  V3,  etc. 


THEOKEM  OF  PYTHAGORAS  225 

PROPOSITION  XI.    THEOREM 

354.  If  a,  b,  and  c  denote  the  sides  of  a  triangle,  and  s  = 
a  +  b  +  c),  the  area  of  the  triangle  is  Vs  (s  —  a)  (s  —  b)  (s  —  c). 


Given  the  triangle  ABC,  the  angle  A  being  acute. 


To  prove  that  the  area  of  A  ABC  —      s  (s  —  a)  (s  —  b)  (s  —  c). 
Proof.    1.  Draw  the  altitude  #D,  and  denote  it  by  h. 
2.  The  area  of 


3.  Now 

4. 

5.  But 


h2  =  c2  -  x2-  =  a2  -  (b  -  x)2. 
c2  +  b2  -  a2 


2b 

=  c2  -x2  = 


a;)  (c  -  a:). 

c2+b2-a 


§333 
Why? 

Why? 
Why? 


_  (2  be  +  c2 '+  &2  -  a2)  (2  be  -  c2  -  b2  +  «2) 
_  (6  +  c  +  a)  (6  +  c  —  a)  (a  —  b  +  c)'(a  +  6  —  c) 
6.  If  a.  +  b  +  c  -  2 s, then &  +  c—  a  =  2s  —  2a  =  2(s  —  a), etc. 


7. 


2  * .  2  (*  —  a) .  2  (*  —  &) .  2  (*  -  c) 


8.  .*.*  =  -  V*(«  —  a)  (*  —  o)(s  —  c). 

9.  Hence  the  area  of 


1.    2 

2^'ft 


Historical  Note.   This  is  known  as  Hero's  formula  for  the  area  of  a  tri- 
angle. Hero  of  Alexandria  (about  75  A. D.)  was  a  famous  Greek  surveyor. 


226 


PLANE  GEOMETRY— BOOK  III 


EXERCISES 

1.  The  sides  of  a  triangle  are  10,  17,  21.    Find  its  area. 
Solution.    Since  s  =  24  ^  A  =  V24  •  14  •  7  ^3 

=  V4 • 6 • 7 • 2 -7-3 
-  V4  •  62  •  72  -  84. 

2.  The  sides  of  a  triangle  are  (a)  26,  35,  51 ;  (b)  75,  176,  229 : 
(c)  104,  111,  175.   Find  the  area  in  each  case. 

3.  Explain  how  the  area  of  any  polygon  may  be  found  by  Propo- 
sition XI. 

4.  Given  the  area  A  of  a  triangle  whose  sides  are  a,  b,  c.    The 

2  A 

altitudes  are  denoted  by  ha,  7/6,  7/c.  respectively.    Prove  that  ha  =  — 

2  A  2  A  a  " 

hh  =  — r~»  hc  = Find  the  altitudes  of  the  A  in  Ex.  1  and  Ex.  2. 

b  c 

5.  A  triangular  park  is  bounded  on  all  sides  by  streets.    Its  sides 
are  208  ft.,  222  ft.,  and  350  ft.    How  far  is  it  from  each  corner  to  the 
opposite  street  ?    Which  of  these  distances  is  the  shortest  ? 

6.  If  two  rods  AB  and  Z?C,  3  and  4  units  long  respectively,  are 
hinged  together  at  B,  and  AB  revolves  about  B,  the  length  of  AC 
(denoted  by  x)  evidently  depends  on  the  size  of  Z.B. 

1.  IiZB  =  90°,  x*  =  32  +  42.    Hence 

2.  If  /.B  <  90°,  x2  =  32  +  42  minus  some  quantity. 

3.  If  Z.B  >  90°  and  <  180°,  x2  =  32  +  42  plus  some  quantity. 

A 


B  4  C        B 

Explain  how  these  relations  enable  us  to  find  out  whether  a  tri- 
angle is  acute,  right,  or  obtuse,  when  we  know  its  sides  (§§231,  232). 

7.  The  sides  of  a  triangle  are  6,  7,  8.  What  kind  of  triangle  is  it? 

8.  Given  four  rods  of  lengths  4,  5,  6,  8.   If  they  are  hinged  together 
at  their  extremities,  three  at  a  time,  how  many  different  triangles 
can  be  made  ?    What  kind  of  triangle  results  in  each  case  ? 

9.  The  base  of  the  gable  triangle  of  a  roof  is  28  ft.    P^ach  of  the 
oblique  rafters  is  17  ft.  long.    What  kind  of  triangle  is  it? 


BOOK  III  227 


IMPORTANT  FORMULAS 

Express  in  words  the  following  formulas: 


Rectangle, 

A  —  b  x  a. 

Square, 

A=l\ 

Kite, 

dxd< 

2 

Parallelogram, 

A=bx  h. 

.  Triangle, 

A=$l»xh 

=  Vs  («  —  a)  (s  —  b)  (s  —  c). 


Equilateral  triangle,  A  =  —  Y3. 

Trapezoid,  A  =  $  h(b  +  b'). 

Circumscribed  polygon,      A  =  %p  x  r. 

r~  d 

Square,  d  =  et*v2  ;  a  =  —= 

V2 

2  A 

Triangle, 


Equilateral  triangle,  h  =  -  V3. 


T,.   ,  ,  ,   .        , 
Right  triangle, 


a  x 


REVIEW  EXERCISER 

1.  Why  is  a  square  the  most  convenient  unit  of  area? 

2.  Why  is  the  Pythagorean  theorem  so  important? 

3.  In  how  many  ways  can  you  find  the  area  of  a  polygon  ? 

4.  What  is  the  importance  of  Proposition  V  ? 

5.  The  dimensions  of  a  rectangle  are  4  in.  and  5  in.    What  is  the 
altitude  of  an  equal  equilateral  triangle  ? 

6.  The  diagonals  of  a  rhombus  are  8  and  7.    What  is  the  side  of 
an  equal  square  ? 


228  PLANE  GEOMETKY— BOOK  III 

7.  One  leg  of  a  right  triangle  is  4.    The  angle  opposite  the  leg  is 
30°.    Find  the  area  of  the  triangle. 

8.  The  bases  of  a  trapezoid  are  13  and  10.  What  is  the  altitude,  if 
the  trapezoid  is  equal  to  an  isosceles  right  triangle  whose  hypotenuse 
is  10? 

9.  The  sides  of  three  squares  are  5,  12,  84  respectively.    How 
large  is  a  side  of  the  square  equal  to  their  sum  ? 

10.  A  triangle  has  sides  of  9,  10,  and  17  cm.    Transform  it  into 
a  square.    How  long  should  a  side  of  the  square  be?   Check  by 
measurement. 

11.  Given  a  square  of  side  2  a.    In  this  square  inscribe  another 
square  by  joining  the  mid-points  of  the  sides  in  order.    In  the  second 
square  inscribe  a  third  square  similarly.    Find  the  sum  of  the  perim- 
eters and  the  sum  of  the  areas  of  the  three  squares. 

12.  The  sides  of  a  triangle  are  20,  34,  42.    Find  the  areas  of  the 
parts  into  which  the  triangle  is  divided  by  the  bisector  of  the  angle 
formed  by  the  first  two  sides.    (Proposition  IV.) 

13.  Transform  a  square  into  an  equilateral  triangle. 
Suggestion.    Transform  the  square  into  a  triangle  having  an  angle 

of  60°,  and  apply  Proposition  IV. 

14.  What  is  the  area  of  a  pentagon  circumscribed  about  a  circle 
whose  radius  is  20  ft.,  if  the  perimeter  of  the  pentagon  is  150  ft.? 

15.  Fold  a  rectangular  sheet  of  paper  from  one     

corner,   as   shown  in  the   diagram.    The   successive 

creases  are  to  be  equally  distant  from  each  other, 
and  the  ends  of  each  crease  equally  distant  from  the 
corner.  Prove  that  the  ratio  of  the  successive  areas 
between  the  creases  is  1  ;3  :  5  :  7,  etc. 

16.  The  edge  of  a  regular  tetrahedron  is  7  cm.    Find  its  entire 
exterior  surface  and  its  altitude.    (See  Problem  12,  p.  223.) 

17.  Draw  a  circle  with  a  radius  of  4  in.    Draw  a  diameter  and 
divide  it  into  eight  equal  parts.    Through  the  points  of  division  draw 
chords  of  the  circle  perpendicular  to  the  diameter.    Find  the  lengths 
of  these  chords  to  two  decimal  places,  and  compute  the  approximate 
area  of  the  circle  by  Simpson's  Rule  (p.  208). 

Suggestion.    To  calculate  the  chords,  use  the  equation  of  the  circle, 
Ex.  23,  p.  219. 


REVIEW  EXERCISES 


229 


18.  An  irregular  piece  of  land  adjoined  a  straight  road.   Its  area 
was  found  by  regarding  the  road  as  the  base  line  and  measuring 
distances  (offsets)  at  right  angles  to  it,  as  in  the  figure.    The  follow- 
ing measurements  resulted  (distances  are  all  from  0)  : 
Distance  (feet)        Offset  (feet) 

0  100 

60  40 

140  110 

260  110 

300 


60 


Find  the  area  of  the  field. 


0 


Road 


19.  At  a  certain  point  in  its  course  a  river  is  100  yd.  wide.    From 
a  bridge  built  across  the  river  at  that  point  soundings  are  made  at 
intervals  of  10  yd.,  from  bank  to  bank.    The  water  is  found  to  have 
the  following  depths  (in  feet)  :  5,  10,  12,  20,  20,  25,  20,  18,  8.    If  the 
water  is  flowing  at  an  average  velocity  of  3  mi.  per  hour,  how  many 
gallons,  approximately,  pass  under  the  bridge  per  second  ? 

Suggestion.  Draw  a  diagram  to  scale,  and  proceed  as  in  Ex.  4,  p.  207. 
Calculate  the  result  also  by  Simpson's  Rule  (p.  208). 

The  three  following  exercises  pertain  to   shapes  used  in  steel- 
construction  work,  the  dimensions  being  in  actual  use. 

20.  Find  the  area  of  the  accompanying  "  bevel  nose "  for  the 
dimensions  given  (inches)  : 


a 

b 

c 

d 

t 

H 

-h 

* 

1 

A 

11 

iV 

A 

n 

\ 

li 

li 

& 

3 

\ 

T 


r 


21.  Find  the  area  of  each  of  the  accompanying  special  bevels : 

»tT 


230 


PLANE  GEOMETRY— BOOK  III 


22.  Verify   the   formulas   given   in   engineering   works    for   the 
following  shapes : 

In  Fig.  1,  area  =  (d  +  2  c)  /.    In  Fig.  2,  area  =  tit  +  (s  +  y)  2  z. 


FIG.  1 


FIG.  3 


23.  In  Fig.  3  is  shown  a  rapid  method  of  constructing  a  regular 
octagon.    A  BCD  is  a  square.    Use  the  vertices  as  centers,  and  dis- 
tances equal  to  A  0  as  radii  for  the  interior  arcs.    Prove  that  the 
octagon  is  equiangular  and  equilateral. 

Suggestion.   Let  A  B  =  2  a.  Then  A  0  =  a  V2,  and  A  F  =  2  a  -  a  V2. 

24.  If  one  side  of  a  regular  octagon  is  a,  prove  that  its  area  is 


Suggestion.    Divide  the  octagon  into  rectangles  and  triangles. 

25.  What  is  the  area  of  the  octagon  in  Ex.  23,  in  terms  of  AB'l 

26.  One  of  the  largest  chimneys  in  the  United  States  rests  on  a 
concrete  octagonal  foundation  40  ft.  wide  and  23  ft.  deep.    How  many 
cubic  yards  of  concrete  were  used  in  the  construction  of  the  founda- 
tion?   (See  Exs.  23-25.) 

27.  A  flagpole  has  a  concrete  octagonal  foundation,  each  side  of 
the  octagon  being  .2^  ft.  long.    How  large  an  area  does  the  foundation 
cover  ? 

28.  A  table  top  has  the  form  of  a  regular  octa- 
gon, each  side  being  1  ft.  long.    Find  its  area. 

29.  The  polishing  drums  in  a  button  factory 
have  the  form  of  prisms  with  regular  octagons  for 
bases.   If  each  side  of  the  octagonal  base  is  10  in. 
and  the  length  of  the  drum  is  3  ft.,  what  is  the 
capacity  of  each  drum  ? 

30.  The  construction  of  the  above  figure  is  apparent.    A  C  and 
EH  are  squares.  If  AB  =  4  a,  and  DE  =  a,  find  the  area  of  the  cross. 


BOOK  IV 

PKOPOKTIONAL  MAGNITUDES.    SIMILAR 
POLYGONS 

355.  Extremes  and  Means.    Fourth  Proportional.    In  the  pro- 
portion a:b  =  c:d, 

the  terms  a  and  d  are  called  the  extremes,  and  the  terms  b  and 
c  are  called  the  means. 

The  fourth  proportional  to  three  given  numbers  is  the  fourth 
term  of  the  proportion  which  has  for  its  first  three  terms  the 
three  given  numbers  taken  in  order. 

Thus  in  the  proportion  4  :  8  =  5  :  10,  10  is  the  fourth  proportional  to 
4,  8,  and  5. 

356.  Continued  Proportion.    The  numbers  a,  b,  c,  d,  are  said  to 
be  in  continued  proportion  if  a  :  b  =  b  :  c  =  c  :  d.    If  three  num- 
bers are  in  continued  proportion,  the  second  is  called  the  mean 
proportional  between  the  other  two,  and  the  third  is  called  the 
third  proportional  to  the  other  two. 

Thus  in  the  proportion  2  :  4  =  4  :  8,  4  is  the  mean  proportional  between 
2  and  8,  and  8  is  the  third  proportional  to  2  and  4. 

357.  Definition  of  Inversely  Proportional.    The  terms  of  one 
ratio  are  said  to  be  inversely  proportional  to  the  terms  of  another 
when  the  first  ratio  is  equal  to  the  reciprocal  of  the  second  ratio. 

Thus  a  and  6  are  said  to  be  inversely  proportional  to  x  and  y,  if 
a  :  b  =  y  :  x.  For  example,  the  two  altitudes  of  a  parallelogram  are  in- 
versely proportional  to  the  corresponding  bases.  If  the  sides  are  a  and 
6  and  the  corresponding  altitudes  ha  and  h^,  then  aha  =  bhb.  (Why  ?) 
Whence,  dividing  both  members  by  the  product  ahb, 

ha  :  hb  =  b  :  a. 

That  is,  the  altitudes  ha  and  h^  of  the  parallelogram  are  inversely  propor- 
tional to  the  corresponding  bases  a  and  6. 

231 


232      PLANE  GEOMETRY— BOOK  IV 

FUNDAMENTAL  PRINCIPLES 

358.  Theorem  I.    In  every  proportion  the  product  of  the  ex- 
tr ernes  is  equal  to  the  product  of  the  means. 

For  if  —  =  — ,  we  may  multiply  the  equals  by  bd.    Then 

O  Ci 

ad  =  be.  This  principle  enables  us  to  find  any  term  of  a  pro- 
portion, if  the  other  three  are  given. 

359.  Theorem  II.    The  mean  proportional  between  two  num- 
bers is  equal  to  the  square  root  of  their  product. 

For  if  a  :  b  =  b  :  c,  then  b2  =  ac.    Therefore  b  =  -Vac. 

.  360.  Theorem  III.  If  the  product  of  a  pair  of  numbers  equals 
the  product  of  a  second  pair,  the  four  numbers  will  be  in  propor- 
tion when  written  in  any  order  that  makes  one  pair  the  extremes 
and  the  other  pair  the  means. 

Thus  if  ad  =  be,  then  it  readily  follows  that 

"1.  a  :  b  =  c  :  d.  5.  b  :  a  =  d  :  c. 

2.  a  :  c  =  b  :  d.  6.  b  :  d  =  a  :  c. 

3.  d  :  b  =  c  :  a.  7.  c  :  a  =  d  :b. 

4.  d  :  c  =  b  :  a.  8.  c  :  d  =  a  :  b. 

361.  Theorem  IV.  In  a  series  of  equal  ratios  the  sum  of  the 
antecedents  is  to  the  sum  of  the  consequents  as  any  antecedent  is 
to  its  consequent. 

Proof.    If  -  =  -  =  -  =  f ,  let  x  represent  the  value  of  each 
o       a      j      li 

of  the  ratios. 

Then  a  =  bx,  c  =  dx}  e  =  fx,  g  =  hx. 

Hence  a  +  c  +  e  +  g=(b  +  d  +/+  h)  x.  Why? 

m,  a  +  c  -\-  e  +  (j  a      c 

Then  ,    '       .    .      ,  =  x  =  -  =  - ,  etc. 

b  +  d  +f+  h  b       d 

1      2      3        ^1+2+3       6       1      2      3 

Thus  -  =  -  =  -,  and  — —  =  —  =  -  =  -  =  -. 

246  2  +  4  +  6      12      2      4      6 


PKOPORTIONAL  MAGNITUDES  233 

TRANSFORMATIONS  OF  PROPORTIONS 

The  properties  of  fractions  and  the  fundamental  principles 
stated  above  lead  at  once  to  these  additional  properties  of 
proportions : 

362.  If  four  numbers  are  in  proportion,  they  are  in  proportion 
by  alternation  ;  that  is,  the  first  term  is  to  the  third  term  as  the 
second  is  to  the  fourth. 

Thus  if  a:b  —  c\d,  then  a-.c  —  b  :  d.  Why? 

363.  If  four  numbers  are  in  proportion,  they  are  in  propor- 
tion by  inversion;  that  is,  the  second  term  is  to  the  first  as  the 
fourth  is  to  the  third. 

Thus  if  a:b  =  c:d,  then  b:a=d:c.  Why? 

364.  If  four  numbers  are  in  proportion,  they  are  in  propor- 
tion by  addition  ;  that  is,  the  sum  of  the  first  two  terms  is  to  the 
second  term  as  the  sum  of  the  last  two  terms  is  to  the  fourth  term. 

Proof.    If  \  =  v  then  |  +  1  =  •£  +  1.  Why  ? 

o       d  o  a 

a  -f  b      c  -f-  d 
Then 


Similarly, 


b  d 

a  +  b      c  -f-  d 


365.  If  four  numbers  are 'in  proportion,  they  are  in  propor- 
tion by  subtraction  ;  that  is,  the  difference  of  the  first  two  terms 
is  to  the  second  term  as  the  difference  of  the  last  two  terms  is  to 
the  fourth  term. 

c     .,        a      „       £  _  . 
d 


Proof.    If 

Then 
Similarly, 

T  =  -  >  then  7  —  1 
b       d             b 

a  —  b      c  —  d 

b             d 
a  —  b      c  —  d 

a              c 

Why? 


234      PLANE  GEOMETRY—  BOOK  IV 

EXERCISES 

1.  In  the  following  proportions  determine  the  value  of  x  : 

(a)  9  :  12  =  8  :  x  ;   (b)  4  :  5  =  12  :  x  ;  (c)  7  :  5  =  x  :  4  ;   (d)  x  :  m  =  c  :  n; 

(e)  =     ;  (f)12:*  =  a::27;  (g)  *  :  a*b  =  ab*  :  x. 


2.  Is  the  following  proportion  correct  :  3^  :  4^  =  5  : 

3.  Given  the  integers  2,  3,  4,  5,  6,  8,  9,  12.    Find  the  fourth  pro- 
portional to  any  three  chosen  at  random;  the  mean  proportional 
between  any  two  ;  the  third  proportional  to  any  two. 

4.  Find  the  third  proportional  to  m  and  n  ;  c  and  d  ;  a  +  b  and  c  +  d. 

5.  Suppose  that,  in  the  annexed  figure,  AB  is  30  in.  long.    Find 
the  lengths  of  A  C,  CD,  and  DB,  MAC:  CD  c         D 

=  2  :  3,  and  CD  :  DB  =  6  :  5.  A  -  1  -  1  -  B 

6.  Write  as  a  proportion  each  of  the  following  equations  :  cd  =  mn  ; 
rs  =  xy  ;  a2  —  b2  =  cd  ;  b2  =  ac  ;  x2  =  5  2/2  ;  p*  =  3  ^2  ;  x2  =  a2  —  ax. 

7.  Two  segments  are  8  ft.  and  12  ft.  long  respectively.    Show 
that  their  ratio  remains  the  same  if  they  are  measured  in  inches  ; 
in  centimeters. 

8.  The  figure  represents  one  form  of  the  straight  lever.  A  straight 
bar  AB  is  supported  at  F  (fulcrum).  A  weight  (W)  is  attached  to  it 
at  B.    Then,  by  exerting  enough       A 


pressure  (P)  at  A,  it  is  possible  to 


keep  the  bar  balanced.  Examples 
of  the  lever  are  furnished  by  the 
ordinary  balance,  the  crowbar,  a 
pair  of  scissors.  Let  the  lengths 
of  AFand  FB  be  m  and  n  respectively.  Numerous  experiments  have 
established  the  law  that  if  the  lever  is  in  equilibrium,  P :  W  =  n  :  m  ; 
that  is,  P  and  W  are  inversely  proportional  to  m  and  n.  For  example, 
if  AF  =  25  in.,  and  FB  =  20  in.,  and  a  weight  of  40  Ib.  is  attached 
at  B,  what  pressure  must  be  applied  at  A  to  raise  the  weight  ? 
Solution.  P  :  40  =  20  :  25. 


.*.  the  pressure  P  is  32  Ib. 
(No  allowance  being  made  for  the  weight  of  the  lever  itself.) 


PROPORTION  235 

In  the  following  table  supply  the  values  of  the  missing  terms : 

P  W  m  n 

40  Ib.  50  Ib.  3  ft.  ? 

35  oz.  70  oz.  ?  10  in. 

50kg.  ?  40cm.  20cm. 

?  44  Ib.  4ft.  8ft. 

9.  Two  weights  of  7  lb.  and  11  Ib.  are  suspended  from  the 
ends  of  a  lever  9  ft.  long.  Find  the  position  of  the  fulcrum  if  the 
lever  is  balanced. 

10.  A  weight  W  is  suspended  first  at  A  and  then  at  B  in  the 
figure  for  Ex.  8,  and  the  corresponding  pressures  Pl  and  P2  are 
found.    Show  that  W  is  a  mean  proportional  between  Pl  and  7J2, 
for  any  fixed  position  of  F. 

11.  The  sides  of  a  triangle  are  7,  8,  and  9.    Find  the  segments  of 
8  made  by  the  bisector  of  the  opposite  angle  (§  338). 

Suggestion.  If  m  and  n  are  the  required  segments,  then  m  :  n  =  7  :  9. 
Hence  m  +  n  :  m  =  16  :  7,  etc. 

12.  Show  that  the  side  of  a  square  is  the   mean  proportional 
between  the  dimensions  of  an  equal  rectangle. 

13.  By  transforming  each  of  the  following  proportions,  test  the 
validity  of  alternation,  inversion,  addition,  subtraction. 

15  : 10  =  9  :  6, 
a:ab  =  b:  b2. 

14.  If  a  :  b  =  c  :  d,  show  that 

a  +  bia  —  b  =  c  +  d:c  —  d. 

In  this  result  a,  &,  c,  and  d  are  said  to  be  in  proportion  by  addition 
and  subtraction. 

15.  Transform  by  addition  and  subtraction  the  proportion 

a  +  b:a  —  b  =  x  +  y  :x  —  y. 

16.  If  b  is  the  mean  proportional  between  a  and  c,  show  that 

a  :  c  =  a?  :  b'2. 

17.  If  a  :  b  =  c  :  d,  show  that 

2  a2  +  3  b2  :  2  a2  -  3  I*  =  2  c2  +  3  d2  :  2  c2  -  3  d2. 

(Let  -  =  -  =  &;  then  a  =  bk,  and  c  =  dk.   Substitute  these  values  of 
b      d 

a  and  c  in  the  given  proportion  and  establish  an  identity.) 


236 


PLANE  GEOMETKY— BOOK  IV 


SIMILAR  POLYGONS 

366.  A  surveyor  is  sent  to  measure  a  tract  of  land,  which 
is  in  the  form  of  an  irregular  polygon,  —  a,  quadrilateral,  for 
example.    He  records  the  number  of  feet  (or  rods)  in  each 
side,  and  the  number  of  degrees  in  each  angle,  or  at  least  in 
as  many  of  the  angles  as  are  neces- 
sary.   He  then  desires  to  represent 

this  tract  of  land  by  means  of  a  draw- 
ing. In  order  to  accomplish  this,  he 
uses  a  scale  of  a  certain  number  of 
feet  (or  rods)  to  the  inch. 

Suppose  that  the  scale  is  20  rd.  to 

the  inch,  and  that  this  figure  is  the  result.  Then  the  ratio 
of  any  side  in  the  diagram  to  the  actual  length  is  1 :  3960 
(=  20  x  161  x  12).  Has  he  made  any  change  in  the  angles  ? 

The  drawing  conveys  a  correct  idea  of  the  tract  of  land, 
for,  though  smaller  in  size,  it  is  of  the  same  shape. 

An  architect,  in  drawing  the  plans  for  a  house,  applies  the 
same  principle. 

367.  If  a  piece  of  cardboard  in  the  shape  of  a  polygon  is  held 
between  a  source  of  light  and  a  wall,  and  parallel  to  the  wall,  a 
shadow  is  cast  which  is  a  copy,  on  a  larger  scale,  of  the  original 
figure. 


What  relation  exists  between  the  angles  of  the  original  figure 
and  those  of  the  shadow  ?  What  relation  between  the  sides  ? 

368.  If  a  polygon  is  formed  by  joining  in  succession 
several  intersections  of  the  heavy  lines  of  squared  paper,  and 
another  figure  is  formed  by  joining  a  corresponding  series  of 


SIMILAR  POLYGONS 


237 


intersections  of  the  lighter  lines,  what  relation  exists  between 
the  angles  of  the  two  polygons  ?  what  relation  between  the 
sides  ? 


Can  the  smaller  figure  be  considered  a  copy  of  the  larger  ? 
If  so,  to  what  scale  ? 

The  proofs  of  the  correct  answers  to  the  above  questions  will  be 
derived  later. 

369.  Mutually  Equiangular  Polygons.    Homologous  Sides  and 
Angles.    Two  polygons  which  have  the  angles  of  one  equal 
respectively  to  the  angles  of  the  other  are  said  to  be  mutually 
equiangular.    In  that  case  two  of  the  equal  angles  (one  from 
each  polygon)  are  known  as  homologous  angles,  and  two  sides 
(one  from  each  polygon)  which  are  similarly  situated  with 
reference  to  the  equal  angles  are  called  homologous  sides. 

370.  Similar  Polygons  are  polygons  which  have  their  angles 
respectively  equal  and  their  homologous  sides  proportional. 

The  similarity  of  two  figures  is  indicated  by  the  symbol  ~ 
(a  horizontal  s).     Thus  P~P' 
means  "P  is  similar  to  P'." 

Similar  figures  have  the  same 
shape. 

If  P  and  P'  are  similar  poly- 
gons, then 


and 


AB        BC 


BC 


CD 


Z.C 


CD 


DE 


B'C'       C'D1      C'D'      D'E1 


238  PLANE  GEOMETRY— BOOK  IV 

371.  Ratio  of  Similitude.  From  the  foregoing  considerations 
it  is  evident  that  any  two  homologous  sides  of  two  similar 
figures  may  be  used  to  determine  the  scale  by  which  one 
figure  may  be  considered  a  copy  of  the  other.  The  ratio  of 
two  homologous  sides  of  two  similar  polygons  is  called  the 
ratio  of  similitude  of  the  polygons. 

EXERCISES 

1.  The  sides  of  a  polygon  are  8,  9, 10, 12,  and  15.    Find  the  sides 
of  a  similar  polygon  if  the  ratio  of  similitude  of  the  two  polygons  is 
1:2  (2:3;  3:4;  4:5;  5:6). 

2.  The  legs  of  a  right  triangle  are  5  and  12,  and  the  hypotenuse 
of  a  similar  right  triangle  is  65.    Find  the  lengths  of  the  legs  of  the 
second  triangle. 

3.  Show  that  two  equilateral  triangles  are  similar. 

4.  Show  that  two  squares  are  similar ;  that  any  two  regular  poly- 
gons of  the  same  number  of  sides  are  similar. 

5.  Construct  a  rectangle  whose  sides  are  4  cm.  and  9  cm.    Con- 
struct a  similar  rectangle,  the  ratio  of  similitude  being  1 :  2. 

6.  Two  rectangular  building  lots  are  of  the  same  shape,  but  each 
dimension  of  one  is  three  times  the  corresponding  dimension  of  the 
other.    Find  the  ratio  of  their  areas ;  of  their  perimeters. 

7.  If  the  ratio  of  similitude  of  two  similar  polygons  is  1,  what  ad- 
ditional relation  exists  between 

them  ?  In  what  sense  is  congru- 
ence a  special  case  of  similarity  ? 
The  adjoining  figures  illus- 
trate the  terms  "similarity," 
"equality,"  and  "congruence," 
as  applied  to  triangles. 

8.  Draw  a  scalene  triangle,  an 
irregular  quadrilateral,  a  penta- 
gon, a  hexagon.    Copy  each  of 

them,  using  in  succession  the  scales  1  : 2,  1 : 3,  2  : 5,  2  : 3.  (Is  measure- 
ment of  angles  necessary  in  each  case?) 


SIMILAR  POLYGONS 


239 


9.  In  drawing  by  measurement  a  polygon  similar  to  a  given 
polygon  (see  Ex.  8),  having  given  a  ratio  of  similitude,  is  it  neces- 
sary to  measure  (or  transfer)  all  the  parts  of  the  given  figure  ?  If 
not,  how  many  parts  may  be  omitted,  and  what  parts  ? 

10.  How  do  two  similar  polygons  compare  in  size,  if  their  ratio 
of  similitude  is  a  proper  fraction  ?  unity  ?  an  improper  fraction  ? 

11.  Given  the  triangle  ABC ;  it  is  required  to  construct  geometri- 
cally another  triangle  similar  to  A  ABC,  which  shall  have  to  A  ABC 
a  ratio  of  similitude  equal  to  3  :  4. 


Note  that  this  may  be  done  in  three  different  ways,  as  in  the  above 
diagrams,  namely,  one  side  and  two  adjoining  angles  of  the  original 
figure  may  be  used,  or  two  sides  and  the  included  angle,  or  three 
sides.  Several  theorems  must  be  established,  however,  before  these 
constructions  can  be  proved. 

12.  A  number  of  straight  railroad  tracks  diverge  from  the  same 
station  at  the  point  0.    Several  trains  leave  the  station  at  the  same 
time,  one  over  each  track,  and  run  at  differ- 
ent rates  of  speed,  say  20,  30,  40,  and  50  mi. 
an  hour.    Suppose  that  the  trains  are  stopped 
simultaneously  at  the  end  of  15  min.,  and 
again  at  the   end  of  30  min.,  etc.    Make   a 
drawing  illustrating  the  statement,  using  a 
scale  of  5  mi.  to  the  inch.    What  relation  is 
there  between  the  figures  formed  by  joining 

in  succession  the  stopping  points  in  each  set.  Test  by  measurement 
with  a  protractor,  or  by  actual  construction,  the  equality  of  any  of 
the  angles. 


240  PLANE  GEOMETRY— BOOK  IV 

PROPORTIONAL   SEGMENTS 
PROPOSITION  I.    THEOREM 

372.  If  a  line  is  drawn  through  two  sides  of  a  triangle, 
parallel  to  the  third  side,  it  divides  those  sides  proportionally. 


B  v 

Given  the  triangle  ABC,  with  DE  parallel  to  BC,  dividing  the  sides 
AB  and  AC  into  the  segments  772  and  n,  and  p  and  q,  respectively. 

To  prove  that  m:n=p:q. 

Proof.    1.  '     Draw  BE  and  CD. 

2.  Then  A  A  DE  :  A  BDE  =  m  :  n,                          §334 
and  AADE:ACDE=p:q.                           §334 

3.  But  ABDE  =  ACDE.                              Why? 

.'.  AADE:  ABDE  =  AADE  :  ACDE.  Ax.  5 

.'.  m  :  n  =  p  :  q.  Ax.  1 

373.  The  above  conclusion  may  be  called  the  "  primary 
sense  "  of  this  theorem.  The  word  "  proportionally,"  however, 
may  be  extended  in  its  meaning  to  include  the  results  obtained 
by  taking  the  above  proportion  (§§  362-364)  : 

by  inversion,  n  :  m  =  q  :  p ; 

by  alternation,  m  :  p  =  n  :  q ; 

by  addition,  m  +  n:m  =p  +  q  :p; 

that  is,  AB  :  AD  =  AC  :  AE. 


PROPORTIONAL  SEGMENTS  241 

This  last  form  of  the  proportion  appears  very  frequently. 
In  general  form  it  is  stated  as  follows : 

374.  COROLLARY  l.Ifa  line  is  drawn  through  two  sides  of  a 
triangle, parallel  to  the  third  side, 

one  of  those  sides  is  to  either  part 
cut  off  by  the  parallel  line  as  the 
other  side  is  to  the  correspond- 
ing part. 

375.  COROLLARY  2.     If  two     ~ 
lines  are  cut  by  any  number  of 
parallels,  the  corresponding  seg-     - 

ments  are  proportional.  /  \ 

Draw  BK  and  DL  parallel  to  AG. 

Then  m'  =  m,  p"  —p'=p^  and  r'  =  r.  .  Why  ? 

But  m' :  n  =  p':  q,  and  p":  q  =  r':  s.  §  372 

/.  m  :  n  =  p  :  q  =  r  :  s. 


PROPOSITION  II.    PROBLEM 
376.    To  construct  the  fourth  proportional  to  three  given  lines. 


\ 


Given  the  lines  m,  n,  and  p. 

Required  to  construct  the  fourth  proportional  to  m,  n,  and  p. 

Construction.    1.   Draw  any  angle  GAH. 

2.  On  AG  take  AB  =  m,  BC  =  n;  on  AH  take  AD  =  p. 

3.  Draw  BD.    Through  C  draw  a  line  parallel  to  BD,  meet- 
ing AH  in  E.  Then  DE  is  the  required  fourth  proportional. 

Proof.  (To  be  completed.) 


242      PLANE  GEOMETRY— BOOK  IV 

PROPOSITION  III.    PROBLEM 

377.   To  divide  a  given  line  into  segments  proportional  to 
any  number  of  given  lines. 

m A  ?  P B 

A"5^-          7  T  -;B 


/ 


p "iSs     , 

^V'a 

Given  the  lines  AB,  TTZ,  n,  and  p. 

Required  to  divide  AB  into  segments  proportional  to  m,  n,  and  p. 

Construction.    1.  Draw  AC,  making  any  convenient   angle 
with  AB. 

(Construction  and  proof  to  be  completed.) 

EXERCISES 

1.  In  how  many  ways  can  the  construction  of  §376  be  made? 
Will  the  result  be  the  same  in  each  case  ?  (Check  by  measurement.) 

2.  Find  a  third  proportional  to  two  given  lines  m  and  n. 

3.  If  a,  b,  c,  d,  and  e  are  five  given  lines,  how  would  you  construct 

bcn        2bcn         be  _        bceT     b  /ce\~]n 

a  line  equal  to  —  ?  to ?  to  —  ?  to  —  -    =  -(  —  )? 

a  a  2  a          ad[_     a\d/\ 

4.  If  a,  b,  c,  and  d  are  four  given  lines,  how  would  you  construct 

a  line  equal  to  -?  to  —  ?  to  —  ? 
a  a  ac 

5.  Two  lines,  a  and  b,  are  the  dimensions  of  a  rectangle.    Con- 
struct the  altitude  of  an  equal  rectangle  whose  base  is  equal  to  a 
third  given  line  c,  without  actually  constructing  the  first  rectangle. 

6.  Construct  the  altitude  of  a  rectangle  of  base  b,  which  is  equal 
to  a  square  of  side  c,  without  actually  constructing  the  square. 

7.  Divide  a  given  square  into  two  rectangles  which  are  to  each 
other  as  two  given  lines  m  and  n. 

8.  If  ra,  n,  p,  q,  and  r  are  the  sides  of  a  polygon,  and  A  G  is  a 
given  line,  show  how  Proposition  III  may  be  applied  to  construct 
the  sides  of  a  polygon  similar  to  the  given  polygon,  with  A  G  as  a 
side  homologous  to  m. 


PROPORTIONAL  SEGMENTS  243 

PROPOSITION  IV.    THEOREM 

378.  If  a  line  divides  two  sides  of  a  triangle  proportionally, 
it  is  parallel  to  the  third  side. 

A 


B  C 

Given  the  triangle  ABC,  and  the  line  DE  drawn  so  that 

AD_  AE 
DB  ~  EC 

To  prove  that  DE  II  BC. 

Proof.    1.  From  the  given  proportion,  by  addition, 

AD  +  DB :  AD  =  AE '+  EC  :  AE, 
or  AB:AD  =  AC:AE.  §364 

2.  Through  D  draw  a  line  DG  II  to  BC,  and  suppose  that  it 
cuts  the  line  A  C  in  a  point  F. 

3.  Then  AB  :  AD  =  AC  :AF.  §372 

4.  But  AB:AD  =  AC:AE; 

.'.AE  =  AF,  Why? 

and  the  points  E  and  F  coincide. 

5.  .'.DE  coincides  with  DF.  Why  ? 
That  is,                              DE  II  BC. 

Discussion.    Why  is  the  addition  form  of  the  proportion  used  in  the 
above  proof  ?    (This  theorem  is  the  converse  of  Proposition  I,  §  372.) 


244       PLANE  GEOMETRY— BOOK  IV 

SIMILAR  TRIANGLES 
PROPOSITION  V.    THEOREM 

379.  If  two  triangles  have  the  angles  of  one  equal  respectively 
to  the  angles  of  the  other,  the  triangles  are  similar. 


C      B'  C" 


Given  the  triangles  ABC  and  A'B'C',  having  the  angles  A,  B,  C 
equal  to  the  angles  A\  B\  C1  respectively. 

Toprove  that  AABC  ~ AA'B'C'. 

Proof.  1.  On  A  B  lay  off  AD  =  A'B',  and  on  AC  lay  off 
AE  =  A'C'.  Draw  DE. 

2.  Then  A  ADE  =  AA'B'C1.  Why? 

3.  Also  DE  II  EC.  Why  ? 

4.  Therefore  AB :  AD  =  AC :  AE ;  Why  ? 
that  is,  AB  :  A  'B'  ==  A  C  :  A  'C'. 

5.  In  like  manner,  by  laying  off  the  corresponding  sides  of 
AA'B'C'  from  B  on  BA  and  EC  it  can  be  shown  that 

AB:A'B'  =  BC:B'C'. 

That  is,  the  homologous  sides  of  A  ABC  and  A'B'C'  are 
proportional. 

6.  .'.AABC~AA'B'C'.  Why? 

380.  COROLLARY  1.  If  two  triangles  have  two  angles  of  one 
equal  respectively  to  two  angles  of  the  other,  the  triangles  are 
similar. 


SIMILAR  TRIANGLES  245 

381.  COROLLARY  2.  If  two   right   triangles    have  an  acute 
angle  of  one  equal  to  an  acute  angle  of  the  other,  the  right 
triangles  are  similar. 

382.  COROLLARY  3.  If  two  isosceles  triangles  have  the  angle 
at  the  vertex,  or  a  base  angle  of  one  equal  to  the  corresponding 
angle  of  the  other,  they  are  similar. 

383.  COROLLARY  4.   Two  equilateral  triangles  are  similar. 

384.  COROLLARY  5.  If  two  triangles  have  their  sides  respec- 
tively parallel,  they  are  similar. 

Produce  one  or  both  of  two  non- 
parallel  sides  until  they  meet.  It  can 
then  be  shown  that  the  triangles  have 
their  angles  respectively  equal. 

385.  COROLLARY  6.  If  two  triangles 
have  their  sides  respectively  perpendicu- 
lar, they  are  similar. 

The  proof  is  similar  to  that  of 
Corollary  5. 

REMARK.  It  can  be  shown  that  in  Corollary  5  and  Corollary  6  the 
parallel  sides  and  the  perpendicular  sides  respectively  are  homologous 
sides. 

EXERCISES 

1.  The  diagonals  of  any  trapezoid  divide  each  other  in  the  same 
ratio. 

2.  A  line  parallel  to  one  side  of  a  triangle,  cutting  the  other  two 
sides,  produced  if  necessary,  forms  a  tri- 
angle similar  to  the  given  triangle. 

3.  In  the  accompanying  figure  ABCD 

is  a  parallelogram.   AB  is  produced  to  E.       £ 

Find  all  the  similar  triangles.  ^ 

4.  ABCD  is  any  quadrilateral  inscribed  in  a  circle.    The  diago- 
nals intersect  at  0.    Prove  that  AAOB~ACOD. 


246  PLANE  GEOMETRY—BOOK  IV 

PROPOSITION  VI.    THEOREM 

386.  If  two  triangles  have  an  angle  of  one  equal  to  an  angle 
of  the  other,  and  the  including  sides  proportional,  the  triangles 
are  similar. 


B'  C" 


Given  the  triangles  ABC  and  A'B'C'  in  which  the  angle  A  is 
equal  to  the  angle  A'  and  AB :  A'B'  =  AC :  A'C'. 

To  prove  that  A  ABC  ~  A  A'B'C'. 

Proof.    1.  On  AB  lay   off  AD  =  A'B',  and   on  AC  lay   off 
AE  =  A'C'.    Draw  DE. 

2.  Then  AADE  =  AA'B'C'.  Why? 


AB  _AC 
AD  ~  AE 


3.  Also  ~  =  ^-  Why? 


4.  Therefore  DE  II  EC.  Why  ? 
Then  Zm  =  Z£, 

and  Zw  =  ZC.  Why? 

5.  .'.  &ADE~&ABC.  Why? 
That  is,  AA'B'C' ~AABC.  Ax.  1 

387.  COROLLARY  1.  If  two  right  triangles  have  the  legs  of 
one  proportional  to  the  legs  of  the  other,  they  are  similar. 

388.  COROLLARY  2.  If  two  right  triangles  have  the  hypote- 
nuse and  a  leg  of  one  proportional  to  the  hypotenuse  and  a  leg 
of  the  other,  they  are  similar. 

Since  it  can  be  shown  by  the  Pythagorean  Theorem  that  the  legs'  of 
the  two  triangles  are  proportional. 


SIMILAR  TRIANGLES 
PROPOSITION  VII.    THEOREM 


247 


389.  If  two  triangles  have  their  sides  respectively  propor- 
tional, they  are  similar. 


C     B' 


C 


Given  the  triangles  ABC  and  A' B'C',  in  which  AB  :  A'B'  = 
AC:A'C'  =  BC:  B'C'. 

To  prove  that  AABC~  AA'B'C'. 

Proof.  1.  On  AB  lay  off  AD  =  A'B',  and  on  AC  lay  off 
AE  =  A'C'.  Draw  DE. 

(It  will  be  proved,  first,  that  AADE^AABC,  and  then 
that  AADE  =  A  A' B'C'). 


2.  Then 
For 

and 

3.  Hence 

4.  But 
and  since 


Hence 

5. 

6. 


AABC~AADE. 

Z.A  is  common, 

AB:AD  =  AC:AE. 

AB  _  BC 
AD~  DE' 
AB        BC  . 
A'B'  ~  B'C'  ' 
AD  =  A'B', 
BC  _   BC 
'  '  DE  ~  B'C'' 
DE  =  B'C'. 


/.  AABC-AA'B'C'. 


§386 

Hyp.  and  Cons. 
§370 

Hyp. 
Cons. 
Ax.  1 

Why? 
s.  s.  s. 
Ax.  1 


NOTE.    Observe  that  in  the  above  demonstration  a  congruence  is  es- 
tablished by  means  of  proportion. 


248 


PLANE  GEOMETRY— BOOK  IV 


390.   Summary.    As  an  aid  in  working  original  problems,  the 
results  found  (§§  379-389)  may  be  stated  as  follows : 


In  order  to  prove 


/four  lines  proportional\ . 


two  angles  equal     / 
Show  that  they  are  homologous  (       .    j  of  similar  triangles. 

In  order  to  prove  two  triangles  similar : 

1.  Show  that  their  angles  are  respectively  equal. 

2.  Show  that  an  angle  of  one  is  equal  to  an  angle  of  the 

other,  with  the  including  sides  proportional. 

3.  Show  that  their  sides  are  respectively  proportional. 

4.  Show  that  they  are  right  triangles  and  have 

(a)  a  pair  of  acute  angles  equal,  or 

(b)  two  pairs  of  corresponding  sides  proportional. 

5.  Show  that  they  are  isosceles  triangles  and  have 

(a)  their  vertex  angles,  ori 

(b)  their  base  angles  j 

6.  Show  that  they  are  both  equilateral  triangles. 

7.  Show  that  their  sides  are  respectively  parallel. 

8.  Show  that  their  sides  are  respectively  perpendicular. 

If  a,  &,  c,  d,  are  four  line-segments,  and  it  is  required  to  prove 
that  ad  =  be,  show  that  a :  b  =  c :  d. 


EXERCISES 

1.  Homologous  altitudes  of 

2.  Homologous  medians  of 

3.  The  radii  of  circles  circumscribed 

about 

4.  The  radii  of  circles  inscribed  in 

5.  Lines     drawn     from     homologous 

vertices  to  the  opposite  sides 
arid  making  equal  angles  with 
homologous  sides  in 


similar  triangles  are  pro- 
portional to  homologous 
sides. 


SIMILAR  TRIANGLES 


249 


6.  A  median  of  a  triangle  bisects  any  line  parallel  to  the  corre- 
sponding base  and  included  between  the  other  two  sides. 

7.  Lines  are  drawn  parallel  to  the  base  of  a  triangle  and  are 
terminated  by  the  other  two  sides.    What  is  the  locus  of   their 
middle  points? 

8.  Parallel  lines  are  drawn  with  their  extremities  in  the  sides  of 
an  angle.    Find  the  locus  of  their  middle  points. 

9.  Given  a  square  A  BCD.   Let  E  be  the  middle  point  of  CD, 
and  draw  BE.  A  line  is  drawn  parallel  to  BE  and  cutting  the  square. 
Let  P  be  the  middle  point  of  the  segment  of  this  line  within  the 
square.    Construct  the  locus  of  P  as  the  line  moves,  always  remain- 
ing parallel  to  BE. 

10.  On  squared  paper  plot  the  points  (1,  2),  (2,  4),  (3,  6).    Prove 
that  these  points  lie  on  a  straight  line. 

11.  If  the  coordinates  of  three  points  are  (a,  &),  (c,  d),  and  (e,f), 

CL         C  6 

and  -  —  -  =  —i  prove  that  the  three  points  lie  on  a  straight  line. 

12.  How  many  pairs  of  similar  triangles  are  formed  when  the 
three  altitudes  of  any  triangle  are  drawn  ? 

13.  Given  that  A  A  DE  and  A  B  Care  similar.  A 
Suppose  &ADE  inverted  and  /.A  replaced  on 

itself.    Then  FG,  the  new  position  of  ED,  is 
said  to  be  antiparallel  to  EC. 

Give  all  the  proportions  which  arise  when 
FG  is  antiparallel  to  BC. 

14.  Show  that  if  the  extremities  of  the  three 
altitudes  of  a  triangle  are  connected  as  in  the 
figure,  three  sets  of  antiparallel  lines  are  ob- 
tained. 

15.  Show  by  the  use  of  antiparallels  that 
the  angles  of  the  smaller  triangle  in  Ex.  14 
(called  the  pedal  triangle)  are  bisected  by 
the  corresponding  altitudes. 

16.  Prove  the  corollary  of  Proposition  IV, 
Book  III,  by  drawing  AE  parallel  to  CD 

(see  the  figure),  to  meet  5  C  produced  in  E.  A        D  B 


\ 


250 


PLANE  GEOMETRY— BOOK  IV 


17.  State  and  prove  the  converse  of  Ex.  16. 

18.  A  number  of  rays  passing  through  the  same  point  intercept 
proportional  parts  on  two  parallel  lines. 

O  A  B      C      D 


E    F      G 


H 


H 


F  E 


19.  In  Ex.  18,  if  OA  :  OE  =  1 : 2,  what  is  the  value  of  OB :  OF? 
of  OC:OG'? 

20.  State  Ex.  19  as  a  locus  problem. 

21.  State  and  prove  the  converse  of  Ex.  18. 

Suggestion.  Let  two  of  the  rays,  say  AE  and  BF,  intersect  at  0. 
Then  show  that  a  line  joining  C  to  0  will  cut  off  on  the  line  EH  a 
distance  from  F  equal  to  FG,  and  will  therefore  pass  through  G  and 
coincide  with  CG. 

22.  The  line  joining  the  mid-points  of  the  bases  of  a  trapezoid 
will,  if  produced,  pass  through  the  point  of  intersection  of  the  non- 
parallel  sides.  j 

23.  ABC  is  a  triangle,  and  0  is  any  point  within 
or  without  the  triangle.   Draw  OA,  OB,  OC.    At  A',  a 
point  in  OA,  draw  A'B'\\  to  AB,  and  draw  B'C'\\  to  EC. 
Pro  ve  that  C^4' 1 1  CM,  and  that  A  .4 '.B'C^  AA.ee.    Ois 

called  the  center  of  similitude  of  the  similar  triangles  ABC  and  A'B'C'. 

24.  Explain  how  the  method  of  Ex.  23  might  be  used  to  copy  a 
triangle  according  to  a  given  scale. 

25.  Given  a  fixed  point  D  within  a  triangle  ABC.    Choose  any 
point  E  on  the  perimeter  of  the  triangle,  draw  DE,  and  let  P  be  the 
middle  point  of  DE.    Construct  the  locus  of  P  as 

E  traces  out  the  whole  perimeter  of  the  triangle. 

26.  By  studying  the  annexed   figure  show 
how  to  inscribe  a  square  in  a  given  triangle. 

(A  square  is  said  to  be  inscribed  in  a  triangle 
when  one  side  of  the  square  lies  wholly  in  one 
side  of  the  triangle,  and  the  other  vertices  of  the  square  lie  in  the 
other  two  sides  of  the  triangle.) 


SIMILAR  TRIANGLES 


251 


NUMERICAL  EXERCISES 

1.  Thales  of  Miletus,  a  Greek  mathematician  (600  B.C.),  is  said 
to  have  computed  the  height  of  pyramids  by  means  of  shadows.    He 
measured  the  length  of  the  shadows  cast  at  the  same  time  by  the 
pyramid  and  by  an  upright  rod  of  known  length.    Explain. 

2.  How  high  is  a  tree  that  casts  a  shadow  60  ft.  long,  if  a  vertical 
post  6  ft.  long  casts  a  shadow  8  ft.  long  at  the  same  time  ? 

3.  Complete  the  following  table  : 


Length  of  rod 

Shadow  of  rod 

Shadow  of  building 

Height  of  building 

3ft. 

5ft. 

60ft. 

4ft.  3  in. 

2ft.  10  in. 

46ft. 

.  6ft.  Sin. 

5ft. 

30ft. 

10ft. 

11  ft.  4  in. 

34ft. 

9ft. 

2ft. 

Oft. 

4.  Similar  triangles  may  be  used  to  measure  distances  indirectly. 
Suggestion.    If  AB  cannot  be  measured  directly, 

draw  BC  ±  to  AB,  and  CD  JL  to  EC.  Place  a  stake 
at  0,  so  that  AOD  is  a  straight  line.  Prove  that 
&OAB~kOCD.  What  measurements  should  be 
taken  ? 

5.  The  moon  is  approximately  240,000  mi.  from  the  earth  and 
its  diameter  is  about  2160  mi.    How  far  from  the  eye  must  a  paper 
disk  1  in.  in  diameter  be  held  to  cover  the  moon's  disk  exactly  ? 

6.  The  centers  of  two  circles  are  10  cm.  apart.  The  radii  of  the  cir- 
cles are  3  cm.  and  2  cm.  respectively.    How  far  from  the  center  of  the 
first  circle  is  the  line  of  centers  cut  by  the  common  interior  tangent? 
by  the  common  exterior  tangent?  How  long  are  the  common  tangents? 

7.  The  diagram  shows  how  squared  paper  may  be 
used  to  divide  a  small  segment  into  a  large  number  of 
equal  parts,  for  example,  10.    If  AB  is  the  given  seg- 
ment, erect  BC.Lto  AB  and  equal  to  10  divisions. 
Draw  A  C. 

draw  FE  I 


At  F,  the  first  division  on   CB  from  C, 
to  BA.    Then  &CEF~AACB.     Prove 


that  EF  is  one  tenth  of  AB.   Similarly,  MN  is  six  tenths  of  AB,  etc. 


252 


PLANE  GEOMETRY— BOOK  IV 


8.  The  method  of  division  explained  in  the  preceding  exercise 
underlies  the  construction  of  the  diagonal  scale.  Explain.  The  scale 
shown  below  has  one  inch  for  its  unit.  What  is  the  length  of 
segments  AB,  CD,  EF,  MF1 


10 


M 


10      8       6       4       2  JS  0  1  2 

9.  Make  a  diagonal  scale  and  with  its  aid  measure  the  hypotenuse 
of  a  right  triangle  whose  legs  are  1  in.  and  2  in. ;  the  diagonal  of  a 
square  of  side  1  in. ;  the  altitude  of  an  equilateral  triangle  of  side  2  in. 
Verify  by  computation. 

10.  The  pantograph  is  a  machine  for  drawing  a  plane  figure  sim- 
ilar to  a  given  plane  figure.  It  was  invented  in  1603  by  Christopher 
Scheiner.  One  form  of  the  instrument  is  shown  in  the  figure. 
AB  and  BC,  DE  and  EF,  are  four  bars  parallel  in  pairs  and  jointed  at 
B,  D,  E,  F.  DEFB  is  a  parallelogram. 

Then,  if is  macfe  equal  to  — — »  the 

AB  BL 

points  A,  E,  and  C  will  always  be  in 
the  same  straight  line,  and  the  ratio 

—  will  always  equal  the  ratio  — —  • 

Now  if  pencils  are  placed  at  E  and  C  while  A  is  a  fixed  pivot, 
the  points  E  and  C  will  describe  similar  figures.  Give  proof.  (Prove 
that  &ADE~&ABC.-) 

The  pantograph  is  widely  used  for  reducing  and  enlarging  maps, 
drawings,  etc. 


SIMILAR  TRIANGLES 
PROPOSITION  VIII.    THEOREM 


253 


391.   The  homologous  altitudes  of  two  similar  triangles  have 
the  same  ratio  as  any  two  homologous  sides. 

C . 

C' 


Given  the  two   similar  triangles  ABC  and  A'B'C',  with  the 
homologous  altitudes  CD  and  CD1. 

CD        AC         AB        EC 
To  prove  that        _  =  _  =  _  =  _. 

Proof.    1.  Rt.  triangles  CD  A  and  C'D'A'  are  similar.   Why? 
(To  be  completed.) 


EXERCISES 


=  4.    Prove  that 


1.  Plot  the  graph  of  the  equation 
it  is  a  straight  line. 

2.  The  annexed  figure  represents  a  pair  of  proportional 
compasses.     Two   equal  rods   are   hinged  together   at  0. 
Prove  that  A  A  OC  ~  A  DOB.    Prove  that  y\x  =  b:a.  How 
may  such  an  instrument  be  used  to  construct  a  triangle 
similar  to  a  given  triangle? 

3.  The  lower  and  upper  bases  of  a  trapezoid*  are  c 
and  d  respectively,  and  the  altitude  is  h.    If  the  legs 
are  extended  until  they  meet,  what  is  the  altitude  of 
each  of  the  triangles  thus  formed? 

4.  The  height  of  a  right  circular  cone  *  is  10  in. 
and  the  diameter  of  its  base  is  5  in.    What  is  the 
diameter  of  the  circular  section  made  by  a  plane  3  in. 
from  the  base  ? 


*  A  right  circular  cone  is  generated  by  revolving  a  right  triangle 
about  one  of  its  legs  as  an  axis. 


254 


PLANE  GEOMETRY— BOOK  IV 
PROPOSITION  IX.    THEOREM 


392.  The  altitude  on  the  hypotenuse  of  a  right  triangle 
divides  the  triangle  into  two  triangles  each  similar  to  the  given 
triangle,  and  hence  similar  to  each  other. 


D 


Why? 


Given  the  right  triangle  ABC,  with  CD  perpendicular  to  the 
hypotenuse  AB. 

To  prove  that        rt.  A  A  CD  ~  rt.  A  A-BC  ~  rt.  A  BCD. 
Proof.  (To  be  completed.) 

If  the  lines  in  the  above  figure  are  given  values  represented 
by  the  small  letters,  it  follows  that 

i.  f  =  * 

h       e, 
Hence 

2. 
Hence 


Hence 

3.  Also 

4.  Also 


h2  =  de. 

(1) 

c       a 
a      d 

Why? 

a*  =  cd. 

(2) 

c  _b 
b      e 

Why? 

b2  =  ce. 

(3) 

ce 


e 
ce 


Dividing  (2)  by  (3). 
Adding  (2)  and  (3). 


SIMILAR  TRIANGLES  255 

Statements  1-3  in  general  form  are  as  follows : 
If  in  a  right  triangle  a  perpendicular  is  let  fall  from  the  ver- 
tex of  the  right  angle  upon  the  hypotenuse  : 

393.  COROLLARY  1.    The  perpendicular  is  the  mean  propor- 
tional between  the  segments  of  the  hypotenuse. 

394.  COROLLARY  2.  Either  leg  is  the  mean  proportional  be- 
tween the  whole  hypotenuse  and  the  adjacent  segment. 

395.  COROLLARY  3.   The  squares  of  the  legs  are  to  each  other 
as  the  adjacent  segments  of  the  hypotenuse. 

The  Pythagorean  Theorem  follows  also  as  a  corollary  (f rom"4). 

396.  COROLLARY  4.   The  square  of  the  hypotenuse  of  a  right 
triangle  is  equal  to  the  sum  of  the  squares  of  the  legs. 

From  Proposition  IX  and  Corollary  1  follows  directly 

397.  COROLLARY  5.  If  a  perpendicular  is  let  fall 
from  any  point  in  a  circle  upon  any  diameter,  it 

is  the  mean  proportional  between  the  segments  of  the  diameter. 


PROPOSITION  X.    PROBLEM 

398.    To  construct  the  mean  proportional  between  two  given 
lines. 


a 


\ 


A  B  G 

Given  the  lines  a  and  b. 

Required  to  construct  the  mean  proportional  between  a  and  b. 

Construction.    1.  Draw  AB  =  a,  and  produce  AB  to  C  so  that 
EC  =  b. 

2.  On  A  C  as  a  diameter  describe  a  semicircle. 

3.  At  B  erect  a  perpendicular  upon  AC,  meeting  the  circle 
in  D.  Then  BD  is  the  required  mean  proportional.  Why  ? 


256 


PLANE  GEOMETRY— BOOK  IV 


EXERCISES 

1.  A  primitive  method  of  determining  the  distance  from  a  given 
point  A  to  an  inaccessible  point  B  on  the  same  Q 

level  was  based  on  the  use  of  an  instrument  that 

may  be  improvised  by  hinging  a  carpenter's  square 

to  the  top  of  a  pole  set  upright  in  the  ground  at  ^^C 

A.    Show  how,  by  measuring  AD  and  AC,  to  find  the  distance  AB. 

2.  If  a  and  b  are  two  given  lines,  how  would  you  construct  vab ? 

VI' 

3.  Show  how  Proposition  X  may  be  used  to  transform  a  rectangle 
into  a  square  ;  a  polygon  into  a  square.  (See 

§§  340,  350.) 

4.  The  accompanying  figure  gives  a  sec- 
ond construction  for  the  mean  proportional 
(AE)    between   AB   and   AC  (see  §394). 
Prove  this  statement. 

5.  To  transform  a  square  into   a  rectangle,  the   sum   of  whose 
base  and  altitude  is  equal  to  a  given  line.    (Describe  a  semicircle  on 
the  given  line  as  a  diameter,  and  at  a  distance  from  the  diameter 
equal  to  one  side  of  the  square  draw  a  line  parallel  to  the  given  line, 
intersecting  the  semicircle.    From  one  of  the  points  of  intersection 
let  fall  a  perpendicular  on  the  diameter.    The  foot  of  the  perpendic- 
ular marks  off  the  dimensions  of  the  rectangle.  Why  ?  When  is  the 
construction  impossible  ?) 


\ 


6.  To  transform  a  given  triangle  into  an  equilateral  triangle. 
(Transform  the  given  triangle  into  a  triangle  one  c 

of  whose  angles  is  60°,  and  find  a  mean  proportional 
between  the  two  sides  inclosing  the  angle.) 

7.  A  door  is  surmounted  by  a  circular  arch.    The 
span  AB  is  3  ft.  (see  figure),  and  the  height  of  the 
crown  CD  is  6  in.   Find  the  radius  of  the  circular  arch. 


\T 

v 


SIMILAR  TRIANGLES 
PROPOSITION  XI.    THEOREM 


257 


399.    The  areas  of  two  similar  triangles  are  to  each  other  as 
the  squares  of  any  two  homologous  sides. 


D 


V 


Given  the  two  similar  triangles  ABC  and  A'B'C', 


To  prove  that 


A  ABC 
AA'B'C' 


AB 


Proof.    1.  Draw  the  altitudes  CD  and  C'D'. 


A.   _i_neii 
3.  But 
4. 

AA'B'C' 
AB 

A'B'  x  C'D' 
CD 

~  A'B' 
AB2 

C'D' 
§391 

Ax.l 

A'B' 

.     AABC 

C'D' 

AB         AB 
.  X    .  .    .  — 

AA'B'C'      A'B'      A'B 


EXERCISES 

1.  Prove  the  above  proposition  by  means  of  §  337. 

2.  The   accompanying   diagram   gives   a   simple   illustration    of 
Proposition  XI.    Explain. 

3.  If  two  similar  triangles  have  a  ratio  of 
similitude  of   2:3   [4:5;  a  :  6],   what   is   the 
ratio  of  their  areas  ? 

4.  If  the  ratio  of  the  areas  of  two  similar 
triangles  is  4  : 25  [2:3;  a  :  6],  what  is  their 
ratio  of  similitude  ? 


258      PLANE  GEOMETRY—  BOOK  IV 

EXERCISES 

1.  The  sides  of  a  triangle  are  13,  14,  15.    Find  the  three  sides  of  a 
similar  triangle  whose  area  is  2^  times  the  area  of  the  given  triangle. 

2.  On  the  altitude  of  a  given  equilateral  triangle  as  a  side  a 
second  equilateral  triangle  is  constructed.    What  is  the  ratio  of  the 
areas  of  the  two  triangles  ? 

3.  Find  the  ratio  of  the  areas  of  two  equilateral  triangles  respec- 
tively inscribed  in  and  circumscribed  about  the  same  circle. 

4.  How  far  from  the  vertex  of  a  triangle  of  altitude  10  must  a  line  be 
drawn  parallel  to  the  base  to  divide  the  triangle  into  two  equal  parts  ? 

5.  To  divide  a  triangle  in  a  given  ratio  by  a  line  parallel  to  a 
given  side. 

A* 


B  C  ^J2_/_^ 

Given  the  A  A  B  C. 

Required  to  divide  the  A  ABC  in  the  ratio  3  :  5,  by  a  line  II  to  BC. 
Analysis.    Suppose  the  construction  completed  by  means  of  the 
line  DE. 

Then  &ADE:  trapezoid  DECS  =  3:5.  Cons. 

Whence  A  ADE  :  A  ABC  =  3  :  8.  By  addition 

But  &ADE:  &ABC  =  dz:b2.  §399 

.-.  d2  :  b2  =  3  :  8. 

Sb*2  36 

That  is,     '  d2  =  — - ,  or  d*  =  ft  •  — • 

O  O 

O    T 

It  is  therefore  necessary  first  to  find  a  line  equal  to  — ;  that  is, 

8 

to  find  a  fourth  proportional  to  8,  3,  and  the  side  ft,  and  then  to  find 
a  mean  proportional  between  this  line  and  the  side  ft. 

6.  Construct  DE  in  Ex.  5,  if  the  given  ratio  is  4  :  5  (2  : 3,  5  :  7). 

7.  To  divide  a  triangle  into  five  equal  parts  by  lines  parallel  to 
the  base. 

8.  At  what  distances  from  the  vertex  should  lines  be  drawn  parallel 
to  the  base  to  divide  a  triangle  of  altitude  m  into  n  equal  parts  ? 


TRIGONOMETRIC  RATIOS 


259 


TRIGONOMETRIC  RATIOS 

400.  Draw  an  acute  angle  XA  Y.  From  various  points  in  AX, 
such  as  BV  By  B8,  etc.,  let  fall  perpendiculars  upon  the  other 
side,  meeting  it  at  Cv  C2,  C3,  etc.  A  series  of  similar  triangles 
is  formed.  (Why  ?  See  the  figure.) 

From  this  construction  it  follows  that 


1. 


2. 


3. 


AB, 

ACl 


AC\ 


AC' 


etc. 


etc. 


etc. 


The  ratios  in  each  group  will  be  equal  for  all  positions  of  the 
perpendicular,  that  is,  the  ratios  will  be  equal  for  any  number 
of  right  triangles  containing  the  angle  A. 

T>/"f 

For  example,  if  /.A  is  30°,  the  ratio  — -  for  any  position  of  the  point  B 
has  the  value  \. 

401.  Sine,  Cosine,  and  Tangent  of  an  Angle.  Every  acute  angle, 
therefore,  has  corresponding  to  it  a  set  of  definite  ratios  whose 
values  may  be  found  by  construct- 
ing any  right  triangle  containing 
the  given  angle. 

Suppose  that  Z.  x  is  such  an 
angle,  and  that  a  right  triangle 
is  constructed  to  contain  it.  Let  & 

1)  be  the  leg  of  the  right  triangle  adjoining  the  given  angle, 
a  the  opposite  leg,  and  c  the  hypotenuse.  Then 

The  ratio  -  is  called  the  sine  of  x  (since). 
c  ' 

The  ratio  -  is  called  the  cosine  of  x  (cos  x). 


The  ratio  —  is  called  the  tangent  of  x  (tan  #). 


260  PLANE  GEOMETRY— BOOK  IV 

TRIGONOMETRIC  TABLE 


Angle 

Sin 

Cos 

Tan 

Angle 

Sin 

Cos 

Tan 

1° 

.017 

.9998 

.017 

45° 

.707 

.707 

1.000 

2° 

.035 

.9994 

.035 

46° 

.719 

.695 

1.036 

3° 

.052 

.9986 

.052 

47° 

.731 

.682 

1.072 

4° 

.070 

.9976 

.070 

48° 

.743 

.669 

.111 

5° 

.087 

.996 

.087 

49° 

.755 

.656 

.150 

6° 

.105 

.995 

.105 

50° 

.766 

.643 

.192 

7° 

.122 

.993 

.123 

51° 

.777 

.629 

.235 

8° 

.139 

.990 

.141 

52° 

.788 

.616 

.280 

9° 

.156 

.988 

.158 

53° 

.799 

.602 

1.327 

10° 

.174 

.985 

.176 

54° 

.809 

.588 

1.376 

11° 

.191 

.982 

.194 

55° 

.819 

.574 

1.428 

12° 

.208 

.978 

.213 

56° 

.829 

.559 

1.483 

13° 

.225 

.974 

.231 

57° 

.839 

.545 

1.540 

14° 

.242 

.970 

.249 

58° 

.848 

.530 

1.600 

15° 

.259 

.966 

.268 

59° 

.857 

.515 

1.664 

16° 

.276 

.961 

.287 

60° 

.866 

.500 

1.732 

17° 

.292 

.956 

.306 

61° 

.875 

.485 

1.804 

18° 

.309 

.951 

.325 

62° 

.883 

.469 

1.881 

19° 

.326 

.946 

.344 

63° 

.891 

.454 

1.963 

20° 

.342 

.940 

.364 

64° 

.899 

.438 

2.050 

21° 

.358 

.934 

.384 

65° 

.906 

.423 

2.144 

22° 

.375 

.927 

.404 

66° 

.914 

.407 

2.246 

23° 

.391 

.921 

.424 

67° 

.921 

.391 

2.356 

24° 

.407 

.914 

.445 

68° 

.927 

.375 

2.475 

25° 

.423 

.906 

.466 

69° 

.934 

.358 

2.605 

26° 

.438 

.899 

.488 

70° 

.940 

.342 

2.747 

27° 

.454 

.891 

.510 

71° 

.946 

.326 

2.904 

28° 

.469 

.883 

.532 

72° 

.951 

.309 

3.078 

29° 

.485 

.875 

.554 

73° 

.956 

.292 

3.271 

30° 

.500 

.866 

.577 

74° 

.961 

.276 

3.487 

31° 

.515 

.857 

.601 

75° 

.966 

.259 

3.732 

32° 

.530 

.848 

.625 

76° 

.970 

.242 

4.011 

33° 

.545 

.839 

.649 

77° 

.974 

.225 

4.331 

34° 

.559 

.829 

.675 

78° 

.978 

.208 

4.705 

35° 

.574 

.819 

.700 

79° 

.982 

.191 

5.145 

36° 

.588 

.809 

.727 

80° 

.985 

.174 

5.671 

37° 

.602 

.799 

.754 

81° 

.988 

.156 

6.314 

38° 

.616 

.788 

.781 

82° 

.990 

.139 

7.115 

39° 

.629 

.777 

.810 

83° 

.993 

.122 

8.144 

40° 

.643 

.766 

.839 

84° 

.995 

.105 

9.514 

41° 

.656 

.755 

.869 

85° 

.996 

.087 

11.43 

42° 

.669 

.743 

.900 

86° 

.9976 

.070 

14.30 

43° 

.682 

.731 

.933 

87° 

.9986 

.052 

19.08 

44° 

.695 

.719 

.966 

88° 

.9994 

.035 

28.64 

45° 

.707 

.707 

1.000 

89° 

.9998 

.017 

57.29 

TRIGONOMETRIC  RATIOS  261 

402.  On  the  opposite  page  is  shown  a  table  which  gives  the 
values  to  three  decimal  places  of  the  sine,  cosine,  and  tangent 
of  all  angles  in  degrees  from  1°  to  89°. 

To  illustrate  the  use  of  the  Table,  find  the  value  of  sin  35°.  In  the  first 
column,  under  Angle,  look  for  35°,  and  opposite  this  in  the  second  column, 
with  the  caption  Sin,  we  find  the  number  .574.  Hence  sin  35°  =  .574. 

The  Table  is  also  used  to  find  the  value  of  an  angle  when  its  sine, 
cosine,  or  tangent  are  given.  For  example,  to  find  the  angle  whose 
tangent  equals  .933,  look  for  this  number  in  one  of  the  columns  with 
the  caption  Tan.  We  find  .933  opposite  43°.  Hence  this  is  the  angle 
sought.  If  the  given  number  lies  between  two  numbers  in  the  Table, 
choose  the  one  nearer  the  given  number,  and  the  corresponding  angle. 
For  example,  for  the  angle  whose  cosine  equals  .432,  choose  64°. 

What  change  takes  place  in  the  value  of  the  sine  as  the  angle 
increases  ?  in  the  value  of  the  cosine  ?  of  the  tangent  ? 


EXERCISES 

1.  On  a  sheet  of  squared  paper  and  with  the  aid  of  a  protractor, 
draw  angles  of  10°,  20°,  30°,  •  •  •,  80°,  and  determine  by  measure- 
ment the  approximate  values  of  the  sine,  cosine,  and  tangent  of 
these  angles,  to  two  decimal  places.    (Use  millimeter  paper  if  pos- 
sible.)  Make  a  table  of  these  values  and  compare  the  results  obtained 
with  those  given  in  the  Table,  p.  260. 

2.  Find  by  the  Table  the  values  of  sin  25°;  cos  36°;  tan  85°;  cos  75°; 
sin  62°;  tan  34°;  sin  42°;  cos  54°;  tan  15°. 

3.  Find  without  the  use  of  the  protractor  or  the  Table  the  values 
of  the  sine,  cosine,  and  tangent  of  30° ;  of  45° ;  of  60°.    Express  the 
results  (in  radical  form)  in  a  table  as  follows : 

sine  cosine  tangent 

30° 
45° 
60° 

Check  the  results  by  reducing  to  the  decimal  form  and  comparing 
with  the  Table,  given  that  V2  =  1.414,  V3  =  1.732,  approximately. 

4.  Notice  that  the  values  of  sine  and  cosine  in  the  Table  are  less 
than  unity.    What  is  the  reason  ? 


262 


PLANE  GEOMETRY— BOOK  IV 


APPLICATIONS  OF  TRIGONOMETRIC  RATIOS 

403.  Since  the  values  of  the  sine,  cosine,  and  tangent  of  an 
angle  may  be  ascertained  from  the  Table,  it  is  possible  to  find 
the  length  of  any  side  of  a  right  triangle,  if  another  side  and 
one  of  the  acute  angles  are  known. 

Thus,  suppose  that  in  the  right  triangle  ABC  the  Z.A  and 
the  side  c  are  known. 

B 


Then,  since      -  =  sin  .4,     .'.  a  =  c  x  sin  A  ; 


and  since 


-  =  cos  A ,     .  * .  b  =  c  x  cos  A . 


Similarly,  it  can  be  shown  that  a  =  b  x  tan  A,  b  —  a  x  tan£, 

a 

c  •=  - — -  >  etc. 
sin^. 

For  example,  suppose  it  is  desired  to  find  the  height  of 
a  building  without  actually  measuring  it.  The  horizontal  dis- 
tance AC  can  readily  be  measured,  and  the  angle  at  A  can 
be  determined  by  a  transit  instrument,  or  by  some  simple  sub- 
stitute for  it  (see  §  123).  Suppose  that 
A  C  is  120  ft.,  and  that  the  angle  at  A 
is  35°.  Now  EC  —  AC  -  tan  A.  From 
the  Table,  tan  ,4  =  tan  35°  =  .700. 
Hence  the  height  of  the  building,  or 
EC,  equals  120  x  .700,  or  84  ft. 

If  the  lengths  of  two  sides  of  the  right  triangle  are  given, 
and  the  value  of  the  sine,  cosine,  or  tangent  of  one  of  the  angles 
is  obtained,  the  nearest  value  in  the  Table  under  the  proper 
heading  gives  the  value  of  the  angle  to  the  nearest  degree. 


TRIGONOMETRIC  RATIOS 


263 


EXERCISES 

In  trigonometric  problems,  when  reference  is  made  to  the  right 
triangle  ABC,  it  is  understood  that  the  Z.  C  is  the  right  angle,  and 
that  the  sides  opposite  the  angles  A,  B,  and  C  are  represented  by 
the  letters  a,  &,  and  c  respectively. 

By  using  the  Table  of  values  of  the  trigonometric  ratios  fill  the 
vacant  spaces  in  the  following  table : 


a 

b 

c 

A 

/>> 

1. 

2. 
3 

15 

24 

28 

42° 
65° 

72° 

4. 

56 

50° 

5 

140 

12° 

6.  A  church  steeple  subtends  an  angle  of  35°  at  a  point  on  level 
ground  90  ft.  away.    How  high  is  the  steeple? 

7.  A  cord  is  stretched  from  the  top  of  a  pole  to  a  point  on  the 
ground  48  ft.  from  the  foot  of  the  pole.   It  makes  an  angle  of  50° 
with  the  horizontal.    Find  the  length  of  the  cord  and  the  height  of 
the  pole. 

8.  A  mariner  finds  that  the  angle  of  elevation  of  the  top  of  a 
cliff  is  16°.    He  knows  from  the  location  of  a  buoy  that  his  distance 
from  the  foot  of  the  cliff  is  half  a  mile.    How  high  is  the  cliff? 

9.  A  ladder  resting  against  the  side  of  a  house  makes  an  angle 
of  24°  with  the  house.    If  it  reaches  a  point  18  ft.  from  the  ground, 
how  long  is  the  ladder  ? 

10.  A  man  standing  47  ft.  from  the  corner  of  a  house,  in  line  with 
the  north  and  south  wall,  finds  that  the  east  and  west  wall  subtends 
an  angle  of  52°.    How  long  is  the  east  and  west  wall  ? 

11.  The  Singer  tower,  in  New  York  City,  is  612  ft.  high.    How 
large  an  angle  does  it  subtend  at  a  point  in  New  York  bay,  1£  mi. 
away  ?    (Assume  that  the  base  of  the  tower  is  at  sea  level.) 

12.  What  is  the  angle  of  the  sun's  altitude  if  a  telegraph  pole 
30  ft.  high  casts  a  shadow  42  ft.  long  ? 


264 


PLANE  GEOMETKY— BOOK  IV 


13.  What  is  the  angle  of  the  sun's  altitude  when  a  man's  shadow 
is  twice  his  own  height? 

14.  Find  the  radius  of  a  parallel  of  latitude 
passing  through  Chicago  (42°  N.  Lat.),  if  the 
radius  of  the  earth  is  taken  as  4000  mi.  ? 

(Note  that  in  the  figure  Z  x  =  /.  y.   Why  ?) 

15.  A  balloon  of  diameter  50  ft.  is  directly 
above  an  observer  and  subtends  a  visual  angle 
of  4°.  What  is  the  height  of  the  balloon  ? 

16.  A  house  30  ft.  wide  has  a  gable  roof  whose  rafters  are  21  ft. 
long.    What  is  the  pitch  of  the  roof  (that 

is,  the  angle  between  a  rafter   and  the 
horizontal)  ? 

17.  What  is  the  area  of  the  gable  in  the 
above   example?    (Find  the   altitude   by 
trigonometry.) 

18.  A  boat  has  sailed  20  mi.  in  a  north- 


easterly direction.    What  distance  east  has  it  made  good? 
distance  north? 


What 


19.  A  hexagonal  table  top  is  cut  from  a  circular  slab  of  wood 
4  ft.  in  diameter.    Find  a  side  of  the  hexagon ;  find  its  perimeter 
and  its  area. 

20.  An  observer  standing  at  the  brink  of  the  Grand  Canon  of  the 
Colorado  finds  that  the  opposite  wall  of  the  canon  subtends  an  angle 
of  17°.    The  horizontal  distance  of  the  wall  from  the  point  of  obser- 
vation is  ascertained  to  be  15,600  ft.    How  deep  is  the  cafion  at  that 
point  ? 

21.  Draw  an  acute  triangle  ABC,  and  draw  the  altitude  on  AB. 
Name  the  sides  opposite  the  angles  A,  B,  and  C  by  the  letters  a,  b, 
and  c  respectively,  and  name  the  altitude  h.    Find  the  value  of  h  in 
terms  of  the  side  a  and  the  angle  B,  and  again  in  terms  of  the  side  b 
and  the  angle  A.    By  equating  the  values  thus  found  prove  that 

a  :  b  =  sin  A  :  sin  B. 


22.  In  the  acute  triangle  ABC,  a  =  75  ft.,  /.A  =  46°,  and  ZB  =  17°. 
Find  the  side  b. 


BOOK  IV  265 

CIRCLES  AND  PROPORTIONAL  LINES 
PROPOSITION  XII.    THEOREM 

404.  If  two  chords  intersect  within  a  circle,  the  product  of  the 
segments  of  one  chord  is  equal  to  the  product  of  the  segments 
of  the  other. 

C, 


Given  in  a  circle  two  chords  AB  and  CD  intersecting  at  E. 
To  prove  that  AE  x  EB  ==  CE  x  ED. 

Proof.    1.  Draw  A  C  and  ED. 

2.  Then  A  A  EC  ~  A  BED.  §  380 
For                                       Z.x  =  Zx'. 

(Each  being  measured  by  £  arc  AD.) 
And  Zy  =  Zyf. 

(Each  being  measured  by  |  arc  CB.) 

3.  /.  AE:ED  =  CE  :  EB,  Why? 
or                                    AE  x  EB  =  CE  x  ED.  §  358 

Discussion.  Another  form  of  statement  of  the  proposition  is :  If  two 
chords  intersect  within  a  circle,  the  segments  of  one  chord  are  inversely  pro- 
portional to  the  segments  of  the  other  (§  357).  One  chord  through  E  is 
a  diameter.  Draw  this  diameter  and  then  express  the  product  'of  the 
segments  of  any  chord  through  E  in  terms  of  the  radius  of  the  circle 
and  the  distance  from  E  to  the  center  of  the  circle. 

405.  By  a  secant  from  an  external  point  to  a  circle  is  meant  the 
segment  of  the  secant  lying  between  the  given  external  point 
and  the  farther  point  of  intersection  of  the  secant  and  the  circle. 

The  segment  between  the  external  point  and  the  nearer  point 
of  intersection  is  called  the  external  segment  of  the  secant. 


266       PLANE  GEOMETRY— BOOK  IV 

PROPOSITION  XIII.    THEOREM 

406.  If  from  a  point  without  a  circle  a  secant  and  a  tan- 
gent are  drawn  to  the  circle,  the  tangent  is  the  mean  propor- 
tional between  the  secant  and  its  external  segment. 

D 


Given  AD  a  tangent  and  AC  a  secant  from  the  point  A  to  the 
circle  BCD. 


To  prove  that 
Proof.    1. 
2.  Then 
For 
and 
3. 


AC  :AD  =  AD:AB. 

Draw  ZX7  and  DB. 

A  ADC  ~  AABD. 

Z.  A  is  common  ; 

ZC  =  Z.ADB. 

.  AC  :AD  =  AD:AB. 


380 


Why? 
Why? 

407.  COROLLARY.  If  from  a  fixed  point  without  a  circle  a 
secant  is  draivn,  the  product  of  the  secant  and  its  external  seg- 
ment is  constant,  whatever  the  direction  in  which  the  secant  is 
drawn. 

For  AC  xAB=AD2. 

But  AD  is  constant  for  any  fixed  position  of  the  point  A. 
.•.  AC  x  AB  is  constant. 

Discussion.  In  what  sense  are  Propositions  XII  and  XIII  and  the 
Corollary  to  Proposition  XIII  special  forms  of  one  general  theorem  ? 
State  such  a  theorem. 

How  may  Proposition  XIII  be  used  to  construct  a  mean  proportional 
between  two  given  lines  ?  Is  this  method  to  be  preferred  to  the  one 
already  given? 


CIRCLES  AND  PROPORTIONAL  LINES        267 

NUMERICAL   EXERCISES 

1.  Draw  a  circle  2  in.  in  diameter.   Locate  within  the  circle  any 
point  not  the  center.    Through  this  point  draw  four  chords  of  the 
circle.    Test  by  measurement  the  equality  of  the  products  of  the 
segments  of  the  chords.    (Proposition  XII.) 

2.  A  point  E  is  at  a  distance  of  3  in.  from  the  center  of  a  circle 
whose  radius  is  5  in.    What  is  the  product  of  the  segments  of  any 
chord  drawn  through  £"? 

3.  Proposition  XIII  makes  it  possible  to  compute  the  distance 
one  can  see  on  the  earth's  surface  from  a  point  of  known  elevation 
(no  allowance  being  made  for  refraction  or  for 

the  condition  of  the  atmosphere).    Explain. 

4.  Assuming  that  the  diameter  of  the  earth 
is  8000  mi.,  how  far  can  a  man  see  from  the  top 
of  a  building  100  ft.  high  ?    500  ft.  high  ?  how 
far  from  the  top  of  a  mountain  4000  ft.  high  ? 

5.  Find  the  length  of  a  tangent  to  a  circle 
from  a  point  15  ft.  from  the  center  if  the  radius 
of  the  circle  is  9  ft. 

6.  Find  the  distance  from  a  point  to  a  circle  (§  268)  if  a  tangent 
from  the  point  to  the  circle  is  12  ft.  long  and  the  radius  of  the  circle 
is  5  ft. 

7.  Find  the  radius  of  a  circle  if  a  tangent  from  a  point  29  ft. 
from  the  center  is  21  ft.  long. 

8.  Find  the  radius  of  a  circle  if  a  tangent  from  a  point  8  ft.  from 
the  circle  is  16  ft.  long. 

9.  Find  general  formulas  for  examples  like  Exs.  5,  7,  and  8. 

408.  Extreme  and  Mean  Ratio.  If  a  line  is  divided  into  two 
segments  such  that  one  segment  is  the  mean  proportional  be- 
tween the  whole  line  and  the  other  segment,  then  the  line  is 
said  to  be  divided  in  extreme  and  mean  ratio. 

That  is,  the  point  C  divides  the  line-segment  AB  in  extreme  and 
mean  ratio  if 

AB:AC  =AC:CB. 


268  PLANE  GEOMETRY— BOOK  IV 

PROPOSITION  XIV.    PROBLEM 
409.   To  divide  a  given  line  in  extreme  and  mean  ratio. 


Then 


Given  the  segment  a. 

Required  to  construct  a  segment  x  such  that 

a  :  x  =  x  :  a  —  x. 
Analysis.    Suppose  that  the  construction  has  been  completed. 

a*=a(a-x),  Why? 

or  x2  =  a2  —  ax  ; 

that  is,  x2  +  ##  ==  «2- 

Completing  the  square  in  the  above  quadratic  equation, 

(a\2  /«\2 

-  )  =  a2  +  I-} ; 

that  is,  (x  +  ^ )  =  «'2  +  ( o ) ' 

This  equation  is  of  the  form 


Hence  x  +  -  is  the  hypotenuse  of  a  right  triangle  the  legs  of 
which  are  a  and  -  >  and  x  is  the  difference  between  the  hypote- 
nuse and  the  shorter  leg  of  such  a  right  triangle. 
(Construction  and  proof  to  be  completed.) 


CIRCLES  AND  PROPORTIONAL  LINES        269 

410.  COROLLARY  1.    The  arithmetical  value  of  x  is  about 
.618  of  the  value  of  a. 

For,  completing  the  solution  of  the  quadratic  equation  given  above,  it 

is  found  that  x  +  -  =  ±  -  Vo. 

2  2 

Whence,  neglecting  the  negative  value  of  the  radical, 

a    /-=      a  a  /  /-      .,  \  -^ 

X  =  2  2'  OTX  =  2^-          *'' 

that  is,  x  =  .618  of  a,  nearly. 

NOTE.  The  segment  x  in  the  above  construction  is  called  the  major 
segment,  and  the  segment  a  —  x  the  minor  segment  of  a. 

411.  COROLLARY  2.    If  a  line  which  has  been  divided  in  ex- 
treme and  mean  ratio  is  increased  at  the  extremity  of  its  minor 
segment  by  a  segment  equal  to  its  major  segment,  the  resulting 
line  is  also  divided  in  extreme  and  mean  ratio. 

For,  if  a:x  =  x:a  —  x, 

then  a  +  x  :  a  =  a  :x.  §  364 

REMARK.  This  construction,  with  the  resulting  ratio,  is  celebrated  in 
geometry  under  the  name  of  the  "Golden  Section."  It  is  used  in  the 
construction  of  certain  regular  polygons  in  Book  V.  The  Golden  Sec- 
tion was  used  by  builders  during  the  Middle  Ages  in  many  of  their 
architectural  designs. 

EXERCISES 

1.  Divide  a  line  6  in.  long  in  extreme  and  mean  ratio.    Find  the 
length  of  the  major  segment  by  measurement  and  by  computation, 
and  compare  the  results. 

2.  The  mean  distances  of  the  earth  and  the  planet  Mars  from 
the  sun  are  92,900,000  mi.  and  141,500,000  mi.  respectively.    Show 
that  when  the  three  bodies  are  (approximately)  in  a  straight  line,  the 
greater  distance  is  divided  nearly  in  extreme  and  mean  ratio. 

3.  The  major  segment  of  a  line  which  has  been  divided  in  ex- 
treme and  mean  ratio  is  3  in.    Find  the  length  of  the  line  by  con- 
struction and  measurement  (see  §  409),  and  also  by  computation, 
and  compare  the  results. 


270       PLANE  GEOMETEY— BOOK  IV 

4.  The  minor  segment  of  a  line  which  has  been  divided  in  ex- 
treme and  mean  ratio  is  2  in.    Find  the  length  of  the  line  by  com- 
putation, and  also  by  construction  and  measurement,  and  compare 
the  results. 

5.  Measure  the  length  and  the  width  of  several  books  of  differ- 
ent sizes,  of  the  surface  of  a  rectangular  table,  of  several  rectangular 
picture  frames,  of  the  schoolroom  floor.  Ascertain  whether  the  Golden 
Section  is  apparently  used  in  their  construction. 

6.  In  the  cathedral  of  Cologne  the  height  of  the  towers  from  the 
ground  to  the  roof  is  to  their  total  height  as  5  :  8,  and  the  height  of 
the  cathedral  is  to  the  height  of  the  towers  as  21  : 34.    What  value 
do  these  ratios  approximate  ? 

7.  To  transform  a  given  square  into  a  rectangle  the  difference 
of  whose  base  and  altitude  is  equal  to  a  given  line.    (On  the  given 
line  as  a  diameter  describe 

a  circle.  At  one  end  of  the 
diameter  erect  a  perpendic- 
ular equal  to  a  side  of  the 
given  square.  From  the  other 
end  of  this  perpendicular 
draw  a  secant  through  the  center  of  the  circle.  What  are  the  dimen- 
sions of  the  rectangle  ?  Is  the  construction  always  possible  ?  Why  ?) 

8.  To  construct  a  circle  which  shall  pass  through  two  given 
points  and  touch  a  given  line. 

9.  If  two  circles  are  tangent  externally,  the  segments  of  a  line 
drawn  through  the  point  of  contact  and  terminating  in  the  circles 
are  proportional  to  the  diameters.    Hence  show  that  the  correspond- 
ing segments  of  two  lines  drawn  through  the  point  of  contact  and 
terminating  in  the  circles  are  proportional. 

10.  If  two  circles  are  tangent  internally,  all  chords  of  the  greater 
circle  drawn  from  the  point  of  contact  are  divided  proportionally  by 
the  smaller  circle. 

11.  Two  circles  touch  at  M.    Through  M  three  lines  are  drawn, 
meeting  one  circle  in  A,  B,  C,  and  the  other  in  D,  E,  F  respec- 
tively.   Prove  that  the  triangles  ABC  and  DEF,  formed  by  joining 
the  corresponding  ends  of  the  lines  in  succession,  are  similar. 


BOOK  IV 


271 


SIMILAR  POLYGONS 
PROPOSITION  XV.    THEOREM 

412.  If  two  polygons  are  similar,  they  are  composed  of  the 
same  number  of  triangles,  similar  each  to  each  and  similarly 
placed. 


ir 


Given  the  similar  polygons  P  and  P',  with  the  diagonals  from 
two  homologous  vertices  A  and  A'. 

To  prove  that  A  T  ~  A  T',   A  U  ~  A  U',    etc. 


Proof.    1. 
For 

and 
2. 

3.  But 

4.  Also 
and 

5. 


A  T  ~  A  T1. 


b  :  V  =  h  :  h', 
I  :  V  =  c  :  c'. 
c:c'  =  h:k'. 

ALr~  AU'. 


§386 
Why? 
Why? 
Why? 
Why? 

Ax.  3 
Why? 
Why? 
Why? 
Why? 


6.  In  like  manner  it  can  be  shown  that  the  other  triangles 
are  similar,  each  to  each. 


272      PLANE  GEOMETRY— BOOK  IV 

PROPOSITION  XVI.    THEOREM 

413.  If  two  polygons  are  composed  of  the  same  number  of 
triangles,  similar  each  to  each  and  similarly  placed,  the  poly- 
gons are  similar. 


c 
P 


D 


Given  the  polygons  Pand  Pf,  with  the  triangle  T  similar  to  the 
triangle  T,  the  triangle  U  similar  to  the  triangle  IF,  etc. 

To  prove  that  P  ~  P'. 

Proof.    1.  Z£  =  Z.B'.                                 Why? 

2.  Also  Zm  =  Zmr,                                Why? 
and  Z  n  =  Z  n'. 

.'.  Z  £C7)  =  Z  fi'C'D'.  Ax.  2 

3.  In  like  manner  the  other  angles  are  proved  to  be  respec- 
tively equal. 

4.  Also  b:V  =  h:  h',  Why  ? 
and                                        h:h'  =  c:  c'.                                 Why  ? 

5.  In  like  manner  the  other  homologous  sides  of  the  poly- 
gons are  proved  to  be  proportional. 

6.  .-.  P~P'. 
(Definition  of  similar  polygons,  §  370.) 


SIMILAR  POLYGONS  273 

PROPOSITION  XVII.    PROBLEM 

414.    Upon  a  given  line  homologous  to  one  side  of  a  given 
polygon,  to  construct  a  polygon  similar  to  the  given  polygon. 

D 
E^^7\  D' 


r 


A  B  A'  B1 

Given  the  line  A'B1  homologous  to  AB  of  the  polygon  ABCDEF. 

Required  to  construct  on  A'B'  a  polygon  similar  to  polygon 
ABCDEF. 

Construction.  (Outline.)  Draw  the  diagonals  from  A .  On^'^B' 
homologous  to  AB  construct  a  triangle  similar  to  A  ABC.  On 
A'C'  homologous  to  AC  construct  a  triangle  similar  to  AACD. 
(See  §  380.)  Repeat  this  process  until  for  each  triangle  of 
the  given  polygon  there  is  a  corresponding  similar  triangle 
in  the  other  figure.  Then  polygon  A  'B ' C'D 'E 'F'  ~  polygon 
ABCDEF.  (Why?) 

EXERCISES 

1.  Let  0  be  any  convenient  point  within  or  without  the  polygon 
ABCDE.    Join    0  to  the   vertices  of  the  polygon.    From  A',  any 
point  in  OA,  draw  A  'B'  II  to  AB,  B'  lying  on  OB.    Similarly,  draw 
B'C'  II  to  BC,  etc.    Prove  that  E'A'  II  EA,  and  that  the  new  polygon 
is  similar  to  ABCDE. 

What  name  is  given  to  the  point  0  with  reference  to  the  two 
polygons  ?  (See  Ex.  23,  p.  250.) 

2.  On  coordinate  paper  mark  the  points   (5,  2),  (3,  6),  (7,  8), 
(10,  5),  (8,  2).    Join  these  points  in  succession,  forming  a  polygon. 
Using  the  point  (0,  0)  as  a  center  of  similitude  (Ex.  23,  p.  250),  and 
any  desired  ratio,  construct  a  polygon  similar  to  the  given  polygon. 


274       PLANE  GEOMETRY— BOOK  IV 

PROPOSITION  XVIII.    THEOREM 

415.    The  perimeters  of  two  similar  polygons  are  to  each 
other  as  any  two  homologous  sides. 


Given  the  two  similar  polygons  Q  and  Q',  with  the  perimeters 
p  and  p'  and  the  homologous  sides  a  and  a1. 

To  prove  that  p  \p*  =  a  :  a', 

a       b 


Proof.    1. 


2. 


Why? 


§361 


(In  a  series  of  equal  ratios  the  sum  of  the  antecedents  is  to  the  sum  of 
the  consequents  as  any  antecedent  is  to  its  consequent.) 


3.  That  is, 


p  :p'  =  a  :  a'. 


EXERCISES 

1.  Two  building  lots  have  the  same  shape.   It  takes  400  ft.  of  fence 
to  inclose  one  of  the  lots.   If  two  homologous  sides  are  respectively 
40  ft.  and  50  ft.,  how  many  feet  of  fence  are  required  to  inclose  the 
other  lot  ? 

2.  A  map  is  drawn  to  the  scale  1 : 1,000,000.    How  many  miles 
are  represented  by  the  perimeter  of  a  square  drawn  on  the  map  with 
a  side  1  in.  long  ? 

3.  The  perimeters  of  two  similar  triangles  are  18  in.  and  15  in. 
One  of  the  altitudes  of  the  first  is  6  in.    What  is  the  length  of  the 
homologous  altitude  of  the  other  triangle  ? 


SIMILAR  POLYGONS  275 

PROPOSITION  XIX.    THEOREM 

416.   The  areas  of  two  similar  polygons  are  to  each  other  as 
the  squares  of  any  two  homologous  sides. 


a 
P 


Given  the  two  similar  polygons  P  and  P',with  the  areas  S  and  Sf 
and  the  homologous  sides  a  and  a'. 

To  prove  that  S  :  S'  =r-  "2  •  "'2 


Proof.  1.  Draw  the  diagonals  from  two  homologous  vertices. 
The  two  similar  polygons  are  now  divided  into  pairs  of  similar 
triangles,  T  and  7",  U  and  U',  etc.  (Why  ?) 


2.  Then 

T  _  h*  _U 

T1  ~  h'2  ~  U1 

_  k2  _     V 
~  k12       V1 

§399 

T        U 

V 

AY      1 

T'       U1 

~  V 

AX.    _L 

3TTckT-|/-»0 

T+U+V 

V 

§  361 

.  xie  nee 

T*  +  U'  +  V 

~  V1' 

s 

v 

That  is, 

S' 

=  V'' 

4.  But 

V 

V' 

a2 
~  a12' 

Why? 

s 

a2 

Ax.  1 

'  '  S' 

a12 

276 


PLANE  GEOMETRY— BOOK  IV 


EXERCISES 

1.  The  areas  of  two  similar  polygons  are  to  each  other  as  121  : 169. 
What  is  the  ratio  of  their  perimeters?  of  their  homologous  diagonals? 
of  the  areas  of  two  corresponding  triangles  formed  by  diagonals 
through  homologous  vertices?   What  is  the  ratio  of  similitude  of 
two  such  triangles  ? 

2.  To  construct  a  hexagon  similar  to  a  given  hexagon  and  having 
an  area  twice  as  great. 

3.  To  construct  a  polygon  similar  to  two  given  similar  polygons 
and  equal  to  their  sum  (or  difference).    Construct  a  line  (§  349) 
whose  square  is  equal  to  the  sum  (or  the  difference)  of  the  squares  of 
homologous  sides  of  the  two  given  polygons.    On  this  line  construct 
(§  414)  a  polygon  similar  to  one  of  the  given  polygons.    Prove  the 
construction. 


NUMERICAL  PROPERTIES  OF  TRIANGLES 

417.  Projection.    The  projection  of  one  line  upon  another  is  the 

segment  of  the  second  line  included  between  the  feet  of  the  per- 
pendiculars let  fall  upon  it  from  the  extremities  of  the  first  line. 


D 


CD  (in  the  last  figure  AD)  is  the  projection  of  AB  upon 
MN.  The  line  upon  which  the  projection  is  taken  is  called 
the  base  line. 

418.  Each  leg  of  a  right  triangle  is  the  projection  of  the 
hypotenuse  upon  that  leg. 

419.  The  projection  of  either  leg  of  an  isosceles  triangle  upon 
the  base  is  equal  to  one  half  the  base. 


NUMERICAL  PROPERTIES  OF  TRIANGLES     277 

PROPOSITION  XX.    THEOREM 

420.  In  any  triangle  the  square  of  the  side  opposite  an 
acute  angle  is  equal  to  the  sum  of  the  squares  of  the  other  two 
sides  diminished  by  twice  the  product  of  one  of  those  sides  and 
the  projection  of  the  other  upon  it. 
A 


FIG.  1 


FIG.  2 


Given,  in  the  triangle  ABC,  that  p  is  the  projection  of  the  side  b 
upon  the  side  a,  and  that  the  angle  C  is  acute. 

To  prove  that  c2  =  a2  +  b2  -  2  ap. 

Proof.    1.  Draw  the  altitude  h  upon  the  side  a. 

2.  Then,  in  Fig.  1,    c2  =  h2  +  (a  -  p)2. 

3.  But  h2  =  b2-iA 

4.  /.  c2  =  b2-p2 

=  a2  +  b2-2  ap. 

5.  Also,  in  Fig.  2,     c2  =  h2  +  (p  -  a)2. 

(To  be  completed.) 

EXERCISES 


-     ap 


Why? 
Why? 


Why? 


1.  What  is  the  projection  of  AB  on  BC,  if  AB  is  10  in.  long 
and  makes  with  BC  an  angle  of  30°?   45°?    60°?   90°? 

2.  Show  that  the  projection  of  AB  upon  CD  is  unchanged  if  AB 
is  moved  toward  or  away  from  CD,  remaining  always  parallel  to  its 
original  position  (see  fig.,  p.  276). 

3.  Show  that  the  projection  of  AB  upon  .BC  is  equal  to  the  prod- 
uct of  AB  and  the  cosine  of  the  angle  B. 

4.  Show  that  in  the  above  proposition  c2  —  a2  +  b2  —  2  ab  cos  C. 


278       PLANE  GEOMETRY— BOOK  IV 

PROPOSITION  XXI.    THEOREM 

421.  In  any  obtuse  triangle  the  square  of  the  side  opposite 
the  obtuse  angle  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sides  increased  by  twice  the  product  of  one  of  those  sides 
and  the  projection  of  the  other  upon  it. 


Given,  in  the  triangle  ABC,  that  p  is  the  projection  of  the  side  6 
upon  the  side  a,  and  that  the  angle  BCA  is  obtuse. 

To  prove  that  c2  =  a2  4-  tf  -f  2  ap. 

Proof.    1.  Draw  the  altitude  h  upon  the  side  a. 

2.  Then  c2  =  A2  +  (a  +  ^)2. 

3.  But  A2  =  &2-jA 

(To  be  completed.) 


Why? 
Why? 


Discussion.  When  the  three  sides  of  a  triangle  are  given,  the  last 
two  propositions  make  it  possible  to  find  the  projection  of  any  side 
upon  any  other  side.  The  altitude  upon  any  side  may  then  be  deter- 
mined. Explain  (see  also  §  354). 

From  Propositions  XX  and  XXI  may  be  derived  a  method  of  ascertain- 
ing from  the  lengths  of  the  three  sides  of  a  triangle  whether  the  triangle 
is  acute,  right,  or  obtuse. 

In  a  right  or  an  obtuse  triangle  the  greatest  side  is  opposite  the  right  or 
the  obtuse  angle.  Why  ?  Hence  if  c  is  the  greatest  side  of  a  triangle,  and 

c2  <  a2  +  62,  the  triangle  is  acute  (Proposition  XX). 
If          c2  =  a2  +  &2,  the  triangle  is  right. 
If  c2  >  a2  +  62,  the  triangle  is  obtuse  (Proposition  XXI). 

In  the  above  proposition  show  that  c2  =  a2  +  &2  +  2  ab  cos  (180°  —  C). 


NUMERICAL  PROPERTIES  OF  TRIANGLES     279 

PROPOSITION  XXII.    THEOREM 

422.  In  any  triangle  the  sum  of  the  squares  of  two  sides  is 
equal  to  twice  the  square  of  half  the  third  side,  increased  by 
twice  the  square  of  the  median  on  that  side. 

a 


Given  the  triangle  ABC  in  which  m  is  the  median  to  the  side  c. 
To  prove  that          a2  +  tf  =  2  ( |  j  +  2  m2. 

Proof.  1.  Draw  CE  perpendicular  to  AB,  and  suppose  that 
E  falls  between  A  and  D.  Let  DE,  the  projection  of  CD  upon 
AB,  be  denoted  by  p. 

2.  Then  in  A  BCD,       a2  =  (0V  m2'+  2  (-j\p.  Why  ? 

3.  Also  in  A  A  CD,        b*  =  (0V  m2  -  2  (j\p.  Why  ? 

4.  .'.  a2  +  b2  =  2(0V  2  m2.  Ax.  2 

423.  COROLLARY.   In  any  triangle  the  difference  of  the  squares 
of  two  sides  is  equal  to  twice  the  product  of  the  third  side  and  the 
projection  of  the  median  on  that  side. 

In  the  above  demonstration  a2  —  62  =  2  cp.  Ax.  3 

424.  From  the  result  of  §  422  the  formula  for  the  length  of 
the  median  mc  upon  any  side  of  a  triangle,  as  c,  may  be  derived, 
namely,  =  2          2-2 


280      PLANE  GEOMETRY— BOOK  IV 

PROPOSITION  XXIII.    THEOREM 

425.  In  any  triangle  the  product  of  two  sides  is  equal  to  the 
square  of  the  bisector  of  the  included  angle,  increased  by  the  prod- 
uct of  the  segments  of  the  third  side  made  by  that  bisector. 


Given  the  triangle  ABC,  the  bisector  t  of  the  angle  C,  and  the  seg- 
ments p  and  q  of  the  side  c  made  by  t. 

To  prove  that  ab  =  t2  +  pq. 

Proof.  1.  Circumscribe  a  circle  about  the  &ABC.  Produce  the 
bisector  CD  to  meet  the  circumference  in  E.  Draw  EB,  and  let 
DE  =  u. 

2.  Then  &ACD~&ECB.  §380 
For                                      Z  A  CD  =  Z  ECB,  Hyp. 

and  ZA  =  ZE.  Why? 

3.  .'.  b:t+  u  =  t:a,  Why? 
or                                                    ab  =  t(t  +  u).  Why? 

That  is,  ab  =  t2  +  tu. 

4.  But  tu=pq.  §404 

.  * .  ab  =  P  +  pq. 

426.  From  the  above  relation  and  from  §  338  the  formula  for  the 
length  of  the  bisector  tc  upon  any  side  of  a  triangle,  as  c,  is  derived 
as  follows : 

Solving  for  £,  %  =  ab  —  pq. 

But  from  §  338,  p  :  q  ==  b  :  a. 

.'.  p  :p  +  q  =  b:  a  +  &,  Why? 

and  q:p  +  q  =  a:a  +  b. 


NUMERICAL  PROPERTIES  OF  TRIANGLES     281 


Since,  in  the  figure,    p  +  q  =  c, 


be 


ao 


Substituting, 


•  '•  P  —  — r~r»  and  Q  =  — T~T* 
a + b  a  +  b 

abc* 


By  the  use  of  the  method  and  notation  of  §  354  it  is  found  that 

9 


"\abs(s  —  c). 

U   T    f 

This  formula  holds  for  the  bisector  of  the  angle  C.    Analogous 
formulas  hold  for  the  bisectors  of  angles  A  and  B. 


PROPOSITION  XXIV.    THEOREM 

427.  In  any  triangle  the  product  of  two  sides  is  equal  to  the 
product  of  the  altitude  upon  the  third  side  by  the  diameter  of 
the  circumscribed  circle. 

a 


Given  d,  the  diameter  CE  of  the  circle  circumscribed  about  the  tri- 
angle ABC}  and  the  altitude  h  upon  the  side  c. 

To  prove  that  ab  =  hd. 

(To  be  completed.) 
Suggestion.    Draw  BE  and  compare  A  ACD  and  ECB. 

428.  From  the  above  relation,  and  from  the  formula  (§  354)  for 
the  altitude  upon  any  side  of  a  triangle,  the  formula  for  the  length 
of  the  radius  of  the  circumscribed  circle  may  be  derived  as  follows : 

_  d  _  ab  _  abc 

~~~ 


282 


PLANE  GEOMETRY— BOOK  IV 


PROPOSITION  XXV.    THEOREM 

429.  In  any  triangle  the  area  is  equal  to   the  product  of 
half  the  perimeter  and  the  radius  of  the  inscribed  circle. 


Given  the  triangle  ABC,  and  the  radius  r  of  the  inscribed  circle. 
To  prove  that          area  A  A  EC  =  \(a  +  b  +  c)  >\ 
(To  be  completed.) 

430.  From  the  above  relation,  and  from  the  formula  for  the  area 
of  a  triangle  (§  354),  may  be  derived  a  formula  for  the  length  of  the 
radius  of  the  circle  inscribed  in  any  triangle,  namely, 

r  =  -  "v  s  (.s-  —  a)  (s  —  ft) 


NUMERICAL  EXERCISES 

1.  The  sides  of  a  triangle  are  13,  14,  15.    Find  the  projection  of 
15  upon  14,  and  find  the  altitude  upon  14. 

2.  In  the  following  table  ascertain  whether  the  triangles  indicated 
are  acute,  right,  or  obtuse. 


a 

3 

7 

5 

10 

13 

21 

6 

4 

9 

12 

24 

15 

28 

c 

0 

8 

11 

26 

18 

35 

3.  The  sides  of  a  triangle  are  6,  7,  and  9.    Find  the  projections  of 
9  and  7  upon  6. 

4.  Two  sides  of  a  triangle  are  10  and  12,  and  they  inclose  an 
angle  of  45°.    Find  the  projection  of  10  on  12,  and  of  12  on  10.    Find 
the  altitude  on  12,  and  the  area  of  the  triangle. 


NUMERICAL  PROPERTIES  OF  TRIANGLES     283 

5.  Two  sides  of  a  triangle  are  7  and  8,  and  they  inclose  an  angle 
of  60°.    Find  the  projection  of  the  side  8  on  the  side  7,  and  the  third 
side  of  the  triangle. 

6.  Two  sides  of  a  triangle  are  10  and  12,  and  they  inclose  an  angle 
of  30°.   Find  the  projection  of  the  side  10  on  the  side  12,  the  altitude, 
and  the  area  of  the  triangle. 

7.  Two  sides  of  a  triangle  are  8  and  9,  and  the  angle  between 
them  is  120°.    Find  the  third  side  and  the  area. 

8.  Two  sides  of  a  triangle  are  18  and  30,  the  included  angle  is 
acute,  and  the  projection  of  the  first  upon  the  second  is  12.   Find  the 
third  side. 

9.  One  side  of  an  acute  triangle  is  20,  and  the  projection   of 
another  side  upon  it  is  10.    What  is  known  about  this  triangle? 
Is  the  triangle  determined  definitely? 

10.  One  side  of  an  acute  triangle  is  8,  and  its  projection  on  another 
side  is  4.    What  is  known   about  this  triangle?    Is  the   triangle 
determined  definitely? 

11.  The  sides  of  a  triangle  are  9,  12,  and  15.    Find  the  three 
altitudes. 

12.  The  sides  of  a  triangle  are  5,  9,  and  10.    What  kind  of  triangle 
is  it  ?    Find  the  three  altitudes. 

13.  The  sides  of  a  triangle  are  14,  16,  18.    Find  the  length  of  the 
median  on  the  side  16. 

14.  The  sides  of  a  triangle  are  9,  10,  and  11.    Find  the  length  of 
the  median  on  the  side  9. 

15.  The  sides  of  a  triangle  are  14,  48,  and  50.    Find  the  area,  the 
altitude  on  the  side  50,  and  the  radius  of  the  circumscribed  circle. 

16.  The  legs  of  a  right  triangle  are  2i  and  28.    Find  the  segments 
of  the  hypotenuse  made  by  the  altitude  upon  it. 

17.  Find  the  altitude  on  the  hypotenuse,  and  the  median  on  the 
hypotenuse  of  the  right  triangle  mentioned  in  Ex.  16. 

18.  The  sides  of  a  triangle  are  8,  26,  and  30.    Find  the  radii  of 
the  circumscribed  and  inscribed  circles. 

19.  The  sides  of  a  triangle  are  6,  7,  and  8.    Find  the  length  of  the 
bisector  of  the  angle  opposite  the  side  7,  terminating  in  this  side. 


284      PLANE  GEOMETRY— BOOK  IV 

EXERCISES 
THEOREMS  AND  Locus  PROBLEMS 

1.  The  difference  of  the  squares  of  two  sides  of  a  triangle  is  equal 
to  the  difference  of  the  squares  of  the  segments  made  by  the  altitude 
upon  the  third  side. 

2.  The  sum  of  the  squares  of  the  four  sides  of  a  parallelogram  is 
equal  to  the  sum  of  the  squares  of  the  diagonals  (§§  204,  422). 

3.  The  sum  of  the  squares  of  the  medians  of  a  triangle  is  equal 
to  three  fourths  the  sum  of  the  squares  of  the  three  sides  (§  422). 

4.  The  sum  of  the  squares  of  the  four  sides  of  any  quadrilateral 
is  equal  to  the  sum  of  the  squares  of  the  diagonals  increased  by 
four  times  the  square  of  the  line  joining  the  mid-points  of   the 
diagonals. 

Suggestion.   Join  the  mid-point  of  one  diagonal  to  the  extremities 
of  the  other,  and  apply  §  422. 

5.  If  three  perpendiculars  upon  the  sides  of  a  triangle  meet  in  a 
point,  the  sum  of  the  squares  of  one  set  of  alternate  segments  of  the 
sides  is  equal  to  the  sum  of  the  squares  of  the  other  set  (§  344). 

6.  Find  the  locus  of  points  whose  distances  from  two  fixed  parallel 
lines  are  in  a  given  ratio. 

7.  Through  a  point  A  on  a  circle  chords  are  drawn.    On  each 
one  of  these  chords  a  point  is  taken  one  third  the  distance  from  A 
to  the  end  of  the  chord.    Find  the  locus  of  these  points. 

8.  Given  the  base  of  a  triangle  in  magnitude  and  position  and 
the  sum  of  the  squares  of  the  other  two  sides.    Find  the  locus  of  the 
vertex  (§  422). 

9.  Given  the  base  of  a  triangle  in  magnitude  and  position  and 
the  difference  of  the  squares  of  the  other  two  sides.    Find  the  locus 
of  the  vertex  (§423). 

10.  Plot  the  locus  of  a  point  if  the  product  of  its  distances  from 
two  perpendicular  lines  is  constant  (§  298). 

11.  The  vertex  A  of  a  rectangle  A  BCD  is  fixed,  and  the  directions 
of  the  sides  AB  and  AD  also  are  fixed.    Plot  the  locus  of  the  vertex 
C  if  the  area  of  the  rectangle  is  constant.    (See  Ex.  10.) 


BOOK  V 

REGULAR  POLYGONS  AND  CIRCLES 
CONSTRUCTION  OF  REGULAR  POLYGONS 

431.  Definitions.  A  polygon  that  is  both  equiangular  and 
equilateral  is  called  a  regular  polygon  (§§57,95). 

A  polygon  whose  sides  are  chords  of  a  circle  is  called  an 
inscribed  polygon  (§  255). 

A  polygon  whose  sides  ,are  tangent  to  a  circle  is  called  a 
circumscribed  polygon  (§  277). 

EXERCISES 

1.  If  a  series  of  equal  chords  are  laid  off  in  succession  on  a  circle, 
what   relation   exists   between   the   arcs   of    those 

chords  ?    between  the  central  angles  of  those  arcs  ? 

2.  What  relation  exists  between  the  angles  formed 
by  the  successive  chords  ?    Give  a  reason  for  your 
answer. 

3.  Suppose  that  an  inscribed  polygon  is  formed 

of  equal  chords  of  a  circle.    Why  would  such  a  polygon  be  regular? 

4.  How  many  degrees  in  the  central  angle  of  a  regular  inscribed 
polygon  of  4  (5,  6,  8,  10,  12,  15,  16)  sides?    State  a  general  formula 
for  the  number  of  degrees  in  the  central  angle  of  a  regular  inscribed 
polygon  of  n  sides. 

5.  Find  the  number  of  degrees  in  each  interior  angle  of  each  of 
the  polygons  mentioned  in  the  preceding  exercise.    What  relation 
exists  between  the  central  angle  and  the  interior  angle  of  a  regular 
inscribed  polygon  ?    Give  proof. 

6.  Which  of  the  angles  referred  to  in  Ex.  4  can  be  constructed 
geometrically  by  methods  already  shown  (§§  156,  158,  188)  ? 

Tabulate  the  results  of  the  last  three  exercises. 

285 


286 


PLANE  GEOMETRY— BOOK  V 


7.  Tangents  are  drawn  to  a  circle  at  the  vertices  of  a  regular  in- 
scribed polygon,  forming  a  circumscribed  polygon. 

A  series  of  triangles  is  formed  by  the  sides  of  the 
two  polygons.  How  are  these  triangles  related  to 
each  other?  Why  are  they  isosceles? 

8.  Show  that  a  circumscribed  polygon  such  as  is 
described  in  the  preceding  exercise  is  regular. 

SUMMARY 

432.  To  inscribe  in  a  circle  a  regular  polygon  of  n  sides, 

360° 
construct  a  central  angle  of  -  —  ?  lay  off  the  arc  of  this  angle 

on  the  circle  n  times,  and  then  draw  the  chords  of  these  arcs. 

433.  An  equilateral  polygon  inscribed  in  a  circle  is  regular. 

434.  The  central  angle  of  a  regular  polygon  is  the  supplement 
of  an  interior  angle  at  a  vertex. 

435.  Tangents  drawn  at  the  vertices  of  a  regular  inscribed 
polygon  form    a    regular  circumscribed  polygon  of  the   same 
number  of  sides. 

436.  If  the  mid-points  of  the  arcs  subtended  by  the  sides  of 
a  regular  inscribed  polygon  are  joined  to  the  adjacent  vertices,  a 
regular  inscribed  polygon  of  double  the  number  of  sides  is  formed. 


437.  If  tangents  are  drawn  at  the  mid-points  of  the  arcs  be- 
tween adjacent  points  of  contact  of  the  sides  of  a  regular  cir- 
cumscribed polygon,  a  regular  circumscribed  polygon  of  double 
the  number  of  sides  is  formed. 


REGULAR  POLYGONS  AND  CIRCLES 

PROPOSITION  I.    PROBLEM 

438.   (a)    To  inscribe  a  square  in  a  given  circle. 
(b)    To  circumscribe  a  square  about  a  given  circle. 


287 


(a)  Given  the  circle  O. 

Required  to  inscribe  a  square  in  the  given  circle. 
Construction.    1.  Draw  two  diameters  AC  and  BD  perpen- 
dicular to  each  other.  §  158 
(Construction  and  proof  to  be  completed.) 

439.  COROLLARY.  By  bisecting  the  arcs  AB,  BC,  etc.  a  regu- 
lar polygon  of  8  sides  may  be  inscribed  in  the  circle ;  and  by 
continuing  the  process  regular  polygons  of  16,  82,  64,  etc.  sides 
may  be  inscribed. 

Similarly,  regular  polygons  of  8, 16,  32,  64,  etc.  sides  may  be 
circumscribed  about  a  given  circle. 

Discussion.  Starting  with  the  inscribed  square,  the  construction  of 
§  436  leads  by  repetition  to  regular  inscribed  polygons  of  4  x  2,  4  x  2  x  2, 
4x2x2x2,  etc.  sides,  that  is,  having  4  x  2M  sides,  n  being  a  positive 
integer.  Since  4  =  22,  the  Corollary  may  be  stated  :  A  regular  polygon 
of  2  x  2n  sides  may  be  inscribed  in  or  circumscribed  about  a  given  circle, 
where  n  is  any  positive  integer. 

REMARK.    To  inscribe  a  regular  polygon  with  a  given  number  of  sides, 

360° 
it  must  be  possible  to  construct  the  central  angle  (§  432)  by  means 

of  the  ruler  and  compasses.   Then  the  regular  polygon  can  be  so  con- 
structed and  is  said  to  be  geometric ;  otherwise  not. 


288  PLANE  GEOMETRY— BOOK  V 

PROPOSITION  II.    PROBLEM 
440.   To  inscribe  a  regular  hexagon  in  a  given  circle. 


(Construction  and  proof  to  be  supplied.) 
Suggestion.   ZAOB  =  1™°  =  60°.     .».  A  AOB  is  equilateral. 

441.  COROLLARY  1.   By  joining  the  alternate  vertices  B,  D,  F, 
of  the  regular  inscribed  hexagon,  an  equilateral  triangle  may  be 
inscribed  in  the  circle. 

442.  COROLLARY  2.    By  bisecting   the  arcs  AB,  BC,  etc.  a 
regular  polygon  of  12  sides  may  be  inscribed  in  the  circle  ;  and 
by  repeating  the  process  regular  polygons  of  24,  48,  etc.  sides 
may  be  inscribed.    In  other  words,  regular  polygons  of  3  x  2n 
sides  may  be  inscribed  in  a  given  circle,  and  similarly,  regular 
polygons  of  3  x  #"  sides  may  be  circumscribed  about  a  given  circle, 
n  being  any  integer,  or  zero. 

EXERCISES 

1.  The  radius  of  a  circle  is  5  (10,  R).    How  long  is  the  perimeter 
of  a  regular  inscribed  hexagon?   of  an  inscribed  square?   of  a  cir- 
cumscribed square  ? 

2.  The  perimeter  of  a  regular  inscribed  hexagon  is  42.    Find  the 
diameter  of  the  circle. 

3.  The  diameter  of  a  circle  is  5  (10,  R).    Find  the  area  of  the 
inscribed  square  ;  of  the  circumscribed  square. 

4.  The  radius  of  a  circle  is  R.    What  is  the  perimeter  of  the 
regular  circumscribed  hexagon? 


REGULAR  POLYGONS  AND  CIRCLES          289 

5.  Analysis  of  the  regular  inscribed  hexagon:  Prove  that 

(a)  Three  of  the  diagonals  are  diameters. 

(b)  The  perimeter  contains  three  pairs  of  parallel  sides. 

(c)  Any  diagonal  which  is  a  diameter  divides  the  hexagon  into 
two  isosceles  trapezoids. 

(d)  Radii  drawn  to  the  alternate  vertices  divide  the  hexagon  into 
three  congruent  rhombuses. 

(e)  The  diagonals  joining  the  alternate  vertices  form  an  equilateral 
triangle  whose  area  equals  one  half  the  area  of  the  hexagon. 

(  f  )  The  diagonals  joining  the  corresponding  extremities  of  a  pair 
of  parallel  sides  of  the  hexagon  form  with  these  sides  a  rectangle. 

6.  The  figure  represents  a  regular  hexagon  with  all  its  diagonals. 
Point  out  the  diameters  ;    the  parallel  lines ;    the  rhombuses ;    the 
rectangles ;  the  equilateral  triangles ;  the  right 

triangles ;  the  isosceles  triangles ;  the  pairs  of 
lines  which  bisect  each  other  at  right  angles ; 
the  kites;  the  equal  triangles  (for  example, 
A  ABM  =  &BMN)  ;  a  six-pointed  star;  another 
regular  hexagon. 

How  many  degrees   in   each    angle   of   the 
figure  ? 

7.  If  the  radius  of  a  circle  is  R,  find  the  area  of  the  inscribed 
equilateral  triangle ;   of  the  circumscribed  equilateral  triangle ;   of 
the  regular  inscribed  hexagon. 

8.  The  area  of  the  inscribed  equilateral  triangle  equals  one  fourth 
the  area  of  the  circumscribed  equilateral  triangle. 

9.  The  area  of  the  regular  inscribed  hexagon  is  the  mean  propor- 
tional between  the  areas  of  the  inscribed  and  the  circumscribed 
equilateral  triangles. 

10.  In  Ex.  6  compare  the  area  of  the  given  hexagon  with  that  of 
any  one  of  the  figures  pointed  out. 

11.  If  squares  are  constructed  outwardly  upon  the  six  sides  of  a 
regular   hexagon,  the   exterior  vertices  of   these    squares   are   the 
vertices  of  a  regular  dodecagon. 

12.  Construct  angles  of  30°,  15°,  7£°,  etc. 

13.  Construct  angles  of  45°,  221°,  37^°,  82^°,  52£°. 


290  PLANE  GEOMETKY— BOOK  V 

PROPOSITION  III.    PROBLEM 
443.   To  inscribe  a  regular  decagon  in  a  given  circle. 


Given  the  circle  O. 

Required  to  inscribe  a  regular  decagon  in  the  given  circle. 

Analysis.  1.    If  AB  is  a  side  of  the  required  decagon,  then 
Z.A  OB  =  ^60-°  =  36°.    Hence  Z  OAB  =  Z.  OBA  =  72°.     Why  ? 

2.  But  72°  =  2  x  36°.    Draw  AC  to  bisect  the  Z.OAB. 
Then  in  the  figure,         Z.x  =  Ay  =  36°. 

3.  This  makes  the  A  ABC  and  AOC  isosceles  ;  that  is  AB  = 
AC  =  CO. 

4  Also  AABC~AAOB.  §382 

5.  /.  OB:AB  =  AB:BC, 

or  OB  :  OC  =  OC  :  BC. 

That  is,  the  radius  is  divided  in  extreme  and  mean  ratio 
(§  408),  and  AB  (==  OC)  is.  the  major  segment  of  the  radius. 
(Construction  and  proof  to  be  completed.) 

444.  COROLLARY  1.    The  construction  of  the  regular  decagon 
makes  possible  the  constmction  of  regular 

inscribed  or  circumscribed  polygons  of  5, 10, 
20,  40  (that  is,  5  X  2n),  sides. 

445.  COROLLARY  2.    To  inscribe  in  a  circle 
a  regular  pentadecagon,  or  polygon  of  fifteen 
sides. 

Construct  a  central  angle  of  24°  by  subtracting 
a  central  angle  of  36°  (§  443)  from  one  of  60°  (§  440). 
The  chord  of  this  angle  is  the  side  of  a  regular  inscribed  pentadecagon. 


REGULAR  POLYGONS  AND  CIRCLES          291 

446.  Historical  Note.  Propositions  I-III  establish  the  fact 
that  a  circle  can  be  divided  into  2  x  2W,  3  x  2M,  5  x  2W,  15  x  2n, 
equal  parts,  n  being  any  positive  integer,  or  zero.  The  corre- 
sponding- regular  polygons  were  the  only  ones  known  at  the 
time  of  Euclid,  and  for  centuries  it  was  believed  that  no  other 
regular  polygons  could  be  constructed  with  ruler  and  compasses 
only.  In  1796,  however,  Gauss,  a  famous  German  mathemati- 
cian, found  that  a  circle  can  be  divided  into  17  equal  parts, 
using  only  the  instruments  named.  He  also  answered  the 
general  question,  What  regular  polygons  can  be  constructed 
geometrically  ?  His  result  was  that  the  number  of  sides  must 
be  a  prime  number  of  the  form  2W+  1  (as  2,  3,  5,  and  17),  or  a 
product  of  different  prime  numbers  of  this  form  (as  15,  51,  and 
85),  or  such  a  prime  number  or  product  multiplied  by  a  power 
of  2  (as  3  x  2n  or  15  x  2W). 


EXERCISES 

1.  The  radius  of  a  circle  is  R.    Find  the  side  .s10  of  the  regular 
inscribed  decagon.  Angm    ^  =  |  (VB  _  x). 

2.  Draw  all  the  diagonals  of  a  regular  inscribed  pentagon.    How 
many  degrees  in  each  of  the  angles  of  the  figure? 

Point    out    isosceles    trapezoids ;    rhombuses ; 
isosceles  triangles. 

3.  In  the  figure  for  Ex.  2,  /.  A  CE  contains 
36°.   Does  that  explain  how  a  regular  pentagon 
may  be  constructed  when  one  of  its  diagonals 
is  given  ? 

4.  The  diagonals  of  a  regular  pentagon  by 

their  points  of  intersection  determine  another  regular  pentagon. 

5.  In  a  regular  pentagon  any  two  diagonals  that  are  not  drawn 
from  the  same  vertex  divide  each  other  in  extreme  and  mean  ratio. 

6.  Construct  angles  of  36°,  18°,  9°,  4J°. 

7.  Divide  a  right  angle  into  five  equal  parts. 


292 


PLANE  GEOMETRY— BOOK  V 


8.  In  the  figure  on  page  291  prove  that  ZBAE  is  trisected. 

9.  In  the  figure  let  AB  (=  EC  -  OD)  represent  the  length  of  a 
side  of  the  regular  inscribed  decagon.   Then  A  C  is 

a  side  s5  of  the  regular  inscribed  pentagon.  Then, 
since  Z  DCB  =  36°,  and  Zy  =  18°,  .'.  Zx  =  18°. 
Hence  DE  =  EB.  .'.  DE  is  half  the  difference 
between  the  radius  and  the  side  of  the  decagon. 
Prove  from  this  relation  and  from  Ex.  1,  p.  291, 

that  if  the  radius  is  R,  s&  =  ^  VlO-2V5. 

10.  Show  that  the  sum  of  the  squares  described  upon  a  side  s10  of 
the  regular  inscribed  decagon  and  a  side  s6  of  the  regular  inscribed 
hexagon  equals  the  square  described  on  a  side  s5  of  the  regular 
inscribed  pentagon  (see  Exs.  1  and  9). 

11.  Exercise  10  suggests  a  short  method  of  con- 
structing a  side  ss  of  a  regular  pentagon,  as  shown 
in  the  figure,  in  which  A  C  and  BD  are  diameters, 
and  A  C  _L  BD.    With  E,  the  mid-point  of  OC,  as  a 
center,  and  with  EB  as  a  radius,  describe  an  arc 
cutting  OA  at  F.    Then  OF  =  slo  (by  Ex.  1,  p.  291), 
BF  =  s6,  and  OB  =  .s6.    (Ptolemy,  150  A.D.) 

12.  The  top  of  a  table  has  the  form  of  a  regular  hexagon  (octagon). 
A  carpenter  is  to  strengthen  each  corner  by  a  triangular  strip  of 
wood  which  fits  exactly  under  the  corner.    What  angles  must  he  give 
to  these  pieces  of  wood  if  each  is  in  the  form  of  an  isosceles  triangle  ? 

13.  Show  that  an  angle  of  1^°  can  be  constructed  by  means  of  the 
ruler  and  compasses. 

14.  If  it  were  possible  to  construct  an  angle  of  1°,  the  protractor 
could  then  be  regarded  as  strictly  geometric  (see  footnote,  p.  84). 
What  other  regular  polygons  could  then  be  constructed  geometrically? 

Historical  Note.  If  any  given  angle  could  be  trisected,  it  would  be 
possible  to  construct  a  protractor  geometrically,  for  an  angle  of  1^° 
can  be  constructed.  The  problem  of  trisecting  an  angle  is  one  of  the 
great  historical  questions  which  contributed  materially  to  the  develop- 
ment of  geometry.  While  a  number  of  angles  can  readily  be  trisected, 
e.g.  90°,  108°,  135°,  180°,  a  general  solution  of  the  problem  is  impossible 
by  the  use  of  the  ruler  and  compasses  alone. 


REGULAR  POLYGONS  AND  CIRCLES         293 

PROPOSITION  IV.    THEOREM 

447.  A  circle  may  be  circumscribed  about,  and  a  circle  may 
be  inscribed  in,  any  regular  polygon. 


- -"B 

Given  a  regular  polygon  ABCDEF. 

(a)  To  prove  that  a  circle  may  be  circumscribed  about  ABCDEF. 

Proof.  1.  It  is  possible  to  construct  a  circle  passing  through 
A ,  B,  and  C.  Let  0  be  its  center,  and  draw  OA,  OB,  OC,  and  OD. 

2.  Then  A  A  OB  =  A  COD.  s.  a.  s. 
For                                      AB  =  CD,  Hyp. 

and  OB  =  OC ;  Const, 

also  Z.ABC  =  Z.BCD,  Why  ? 

and  Z OBC  =  Z.OCB,  Why  ? 

whence  /.ABO  =  /.DC O.  Ax.  3 

3.  .\OA  =  OD.  Why? 

.'.  the  circle  passing  through  A,  B,  C,  passes  through  D. 
In  like  manner  it  may  be  proved  that  this  circle  passes 
through  E  and  F. 

(b)   To  prove  that  a  circle  may  be  inscribed  in  ABCDEF. 

Proof.  The  sides  of  the  regular  polygon  are  equal  chords  of 
the  circumscribed  circle.  Hence  they  are  equally  distant  from 
the  center.  §  264 

(To  be  completed.) 


294  PLANE  GEOMETRY—  BOOK  V 

448.  The  center  of  a  regular  polygon  is  the  common  center 
of  the  circumscribed  and  inscribed  circles. 

449.  The  radius  of  a  regular  polygon  is  the  radius  of  the 
circumscribed  circle. 

450.  The  apothem  of  a  regular  polygon  is  the  radius  of  the 
inscribed  circle. 

EXERCISES 

If  R  denotes  the  radius  of  a  regular  inscribed  polygon,  r  the 
apothem,  s  one  side,  A  an  interior  angle,  and  C  the  angle  at  the 
center,  show  that 

1.  In  a  regular  inscribed  triangle  s  =  R  A/3,    r  =  ^R,  A  =  60°, 
C  =  120°. 

2.  In  an  inscribed  square  s  =  R^2,r  =  $R  V2,  A  =  90°,  C  =  90°. 

3.  In  a  regular  inscribed  hexagon  s  =  R,  r  —  \  R  V8,  A  =  120°, 
C  =  60°. 

4.  In  a  regular  inscribed  decagon  (see  Ex.  1,  p.  291) 


5  -  1),  r  =  1  72      io  +  2  V,5,  A  =  144°,  C  =  36°. 

5.  Show  that  the  perimeter  of  a  regular  circumscribed  polygon  is 
greater  than  the  perimeter  of  the  regular  inscribed  polygon  of  the 
same  number  of  sides  (see  §  435). 

6.  Show  that  the  perimeters  and  the  areas  of  regular  inscribed 
polygons  of  4   (8,  16,  32,  etc.)  sides  form  a  series  of  increasing 
numbers. 

7.  Show  that  the  perimeters  and  the  areas  of  regular  circumscribed 
polygons  of  4  (8,  16,  32,  etc.)  sides  form  a  series  of  decreasing  numbers. 

8.  Show  that  the  perimeter  of  a  regular  circumscribed  polygon  of 
4  (8,  16,  32,  etc.)  sides  is  greater  than  the  perimeter  of  the  inscribed 
square.    (Use  Exs.  5  and  6  and  Ax.  12.) 

9.  Show  that  the  perimeter  of  a  regular  inscribed  polygon  of 
4  (8,  16,  32,  etc.)  sides  is  less  than  the  perimeter  of  the  circumscribed 
square    (see  Exs.  5  and  7). 

10.  Prove  Exs.  6-9  if  the  first  polygon  of  the  series  is  one  of  n  sides 
(n  different  from  4)  and  the  fixed  polygon  (Exs.  8  and  9)  also  has 
n  sides. 


REGULAR  POLYGONS  AND  CIRCLES         295 

PROPOSITION  V.    THEOREM 

451.  Two  regular  polygons  of  the  same  number  of  sides  are 
similar. 


Q' 


Given  two  regular  polygons,  Q  and  @f,  each  having  n  sides. 
To  prove  that  Q  and  Q'  are  similar. 

Proof.  (To  be  completed.) 

Suggestion.    1.  Are  the  sides  of  the  polygons  proportional  ? 
2.  Are  the  polygons  mutually  equiangular  ? 

452.  COROLLARY  1.   The  perimeters  of  two  regular  polygons 
of  the  same  number  of  sides  are  to  each  other  as  any  two  homol- 
ogous sides  ;  and  the  areas  of  two  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  the  squares  of  any  two 
homologous  sides.  §§  415,  416 

453.  COROLLARY  2.  In  a  given  circle  a  regular  polygon  may 
be  inscribed  which  is  similar  to  any  given  regular  polygon. 

Suggestion.  Construct  in  the  given  circle  a  central  angle  equal  to  that 
of  the  given  polygon. 

EXERCISES 

1.  On  a  side  of  given  length  to  construct  a  regular  polygon  of  3 
(4,  5,  6,  8,  10,  12,  15)  sides. 

Suggestion.  In  an  arbitrary  circle  with  center  0  construct  a  regular 
polygon  of  the  required  number  of  sides.  Let  AB  be  one  side.  On 
the  given  side  A'B'  homologous  to  AB,  construct  A  A'0'B'~  A  AOB. 
Then  0'  is  the  center  of  the  new  polygon. 

2.  Through  a  given  point  on  a  given  circle  draw  a  chord  so  as  to 
divide  the  circle  into  two  arcs  having  the  ratio  1:3  (1:5,  2:3, 
3  :  7,  7  :  8). 


296  PLANE  GEOMETRY— BOOK  V 

PROPOSITION  VI.    THEOREM 

454.  The  perimeters  of  two  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  their  radii  and  also  as 
their  apothems. 


E 


B 


Given  two  regular  polygons,  each  having  n  sides,  and  with  the 
perimeters  p  and  p\  centers  0  and  O1,  radii  R  and  R,  and  apothems 
randr'. 

To  prove  that  p  :  p'  =  R  :  R\ 

and  p  :p'  =  r  :  r'. 

Proof.    1.  Draw  OB  and  O'B'. 

2.  Then  AAOB  ~  AA'O'B'.                                 §382 
For  A  A  OB  and  A' O'B  are  isosceles,             Why  ? 

and  Z  AOB  =  Z.  A' O'B'. 

(S60°  \ 
Each  being  equal  to ) 
n     ) 

3.  .'.  AB:A'B'  =  R:  R'.  Why? 
Also                     AB:A'B'  =  r:r'.  §391 

4.  But  p:p'  =  AB:A'B'.  §452 

5.  .'.p:p'  =  R:R',  Ax.  1 
and  also                          p  :p' :  :  r  :  r'. 

455.  COROLLARY.  The  areas  of  two  regular  polygons  of  the 
same  number  of  sides  are  to  each  other  as  the  squares  of  their 
radii  and  also  as  the  squares  of  their  apothems.  §  452 


EEGULAR  POLYGONS  AND  CIRCLES          297 

PROPOSITION  VII.    THEOREM 

456.  The  area  of  a  regular  polygon  is  equal  to  half  the 
product  of  its  apothem  and  its  perimeter. 


A      M 

Given  a  regular  polygon,  with  the  perimeter  />,  the  apothem  r, 
and  the  area  S. 

To  prove  that  S  =  \p  x  r. 

Proof.    (Outline.) 

1.  Draw  OA,  OB,  OC,  etc.,  thus  dividing  the  polygon  into 
triangles.    These  triangles  are  congruent.  Why  ? 

2.  Now           area  of  AAOB  =  ^ABxr,  Why  ? 

area  of  ABOC  =  *  BC  x  r,  etc. 

3.  .\  S=i(Xfl  +  fiC  +  ...)xr    Ax.  2 
or  S  =  \p  x  r. 

EXERCISES 

1.  If  a  is  a  side  of  a  regular  hexagon,  find  its  area. 

2.  If  r  is  the  apothem  of  a  regular  hexagon,  find  the  area. 

3.  The  area  of  the  regular  inscribed  octagon  is  equal  to  the  product 
of  a  side  of  the  inscribed  square  and  the  diameter  of  the  circum- 
scribed circle. 

Solution.  Draw  radii  to  three  consecutive  vertices.  Observe  that 
the  octagon  can  be  divided  into  four. equal  kites,  each  having  for  its 
diagonals  one  side  of  the  inscribed  square  and  the  radius  (§  327). 

4.  Prove  that  the  area  of  the  regular  inscribed  dodecagon  is  equal 
to  three  times  the  square  of  the  radius  of  the  circumscribed  circle. 
(Use  the  method  of  Ex.  3.) 


298  PLANE  GEOMETRY— BOOK  V 

CIRCUMFERENCE  OF  A  CIRCLE 
PRELIMINARY  DISCUSSION 

457.  The   following  exercises  will  serve  to  introduce  the 
topic  to  be  discussed : 

1.  Cut  out  of  cardboard  a  circular  disk  4  in.  in  diameter 
and  wind  a  thread  once  around  its  edge.    Measure  the  length 
of  this  thread  in  inches  and  divide  the  number  thus  obtained 
by  the  number  of  inches  in  the  diameter. 

2.  Repeat  Ex.  1,  using  disks  the  diameters  of  which  are 
6,  8,  and  10  in.    Tabulate  the  results  of  Exercises  1  and  2  as 
follows : 


Diameter  =  D 

Length  of  Thread  =  C 

Ratio  C  :  I) 

4 

6 

8 

10 

NOTE.  Measure  to  tenths  of  an  inch  and  compute  to  two  decimal 
places. 

What  do  you  observe  about  the  ratio  C  :  D  ? 

458.  Length  of  the  Circle.  Hitherto  no  reference  has  been 
made  to  the  "  length  of  a  circle "  in  linear  units,  for  the 
reason  that  such  a  term  could  have  no  meaning;  it  would 
be  impossible  to  apply  a  unit  straight  line  to  a  curved  line 
for  purposes  of  measurement.  The  expression  "  length  of  a 
circle "  calls  for  a  definition  all  its  own.  Such  a  definition 
will  be  given  in  a  later  section  (§  487).  For  the  present,  as 
is  illustrated  in  the  exercises  of  the  preceding  section,  by  the 
"  length  of  a  circle  "  is  meant  the  length  of  the  straight  line 
which  would  be  obtained  if  the  circle  were  broken  at  some 
point  and  "  straightened  out."  This  form  of  statement  may  be 
regarded  as  providing  a  practical  definition  of  "  circumference." 


CIRCUMFERENCE  OF  A  CIRCLE  299 

459.  The  circumference  of  a  circle  is  its  length  in  linear  units. 

The  attempt  to  construct,  by  geometric  methods,  a  straight  line  which 
would  be  equal  in  length  to  the  circumference  of  a  circle  of  given  radius 
has  occupied  the  attention  of  geometers  from  early  times.  It  is  now  known 
that  the  solution  of  this  problem  is  not  within  the  means  afforded  by 
elementary  geometry,  and  that  the  numerical  ratio  of  a  circumference 
to  its  diameter  cannot  be  obtained  exactly  by  algebraic  processes. 

The  method  of  §  457  is  not  accurate ;  nor,  if  it  were,  would 
it  be  geometric.  But  an  approximation  to  the  length  of  a  circle 
of  given  diameter  may  be  made  by  a  method  which  is  accurate 
and  is  based  upon  geometric  principles. 

460.  Inscribe  any  convenient  regular  polygon,  say  a  hexagon, 
in  a  circle.    Bisect  each  subtended  arc,  and  join  each  point  of 
bisection  to  the  two  adjoining  vertices  of 

the  polygon.    A  regular  inscribed  polygon 

of  double  the  original  number  of  sides  is 

obtained.    Repeat  the  process,  forming  a 

regular  inscribed  polygon  of  quadruple  the 

original    number  of  sides.    Suppose    now 

that  this  process  of  doubling  the  number 

of  sides  of  the  regular  inscribed  polygon 

were  repeated  several  times.  It  may  be  assumed  for  the  present 

that  the  lengths  of  the  perimeters  of  the  successive  inscribed 

polygons    thus   formed    approach    nearer   and    nearer   to   the 

circumference  of  the  circle,  as  the  number  of  their  sides  is 

increased.    By  computing  these  perimeters  in  succession,  it 

is  possible  to  calculate  the  circumference  of  the  circle  to  any 

desired  degree  of  approximation;  that  is,  to  obtain  the  length 

of  the  circle  to  any  required  degree  of  accuracy. 

To  approximate  to  the  length  of  a  circle,  therefore,  it  is 
necessary  to  compute  the  perimeters  of  successive  regular 
inscribed  polygons  each  of  which  has  twice  as  many  sides 
as  the  preceding  polygon.  The  practical  difficulties  in  mak- 
ing this  calculation  are  greatly  lessened  by  the  solution"  of 
the  following  problem  : 


300  PLANE  GEOMETRY— BOOK  V 

PROPOSITION  VIII.    PROBLEM 

461.  Criven  the  diameter  of  a  circle  and  the  side  of  a 
regular  inscribed  polygon,  to  derive  a  formula  for  the  side  of 
the  regular  inscribed  polygon  of  double  the  number  of  sides. 


Given  CE,  the  diameter  of  a  circle,  AB,  a  side  of  a  regular  inscribed 
polygon,  and  CE  perpendicular  to  AB. 

Required  to  derive  a  formula  for  the  side  AC  of  a  regular 
inscribed  polygon  of  double  the  number  of  sides. 

Solution.    1.  Denote  AB  by  a,  CE  by  d,  A  C  by  x,  and  CD  by  y. 
Then  AD  =  ^  (why  ?),  and  DE  =  d  —  y.    Draw  .4£. 

2.  Then  x2  =  dy.  §394 

3.  But  ^  =  y(d-y)  =  dy-  y\  §  393 

a'" 
Transposing,  y*  —  dy  -\-  —  =  0. 

The  solution  of  this  quadratic  equation  gives  as  the  value  of  y, 


y  = 

4.  But  in  the  case  of  a  regular  polygon  y  is  less  than  \d. 
The  negative  sign  before  the  radical  therefore  gives  the  cor- 
rect value. 


Hence,  from  Step  2, 


x  = 


CIRCUMFERENCE  OF  A  CIRCLE 


301 


462.  COROLLARY.    If  d  —  1,  that  is,  if  a  circle  of  unit  di- 
ameter is  chosen, 


x  = 


-  Vl  -  a2 


or,  also,  x  =  \  V  2  -  2  Vl  -  a2. 

463.  The  results  of  computing  the  perimeters  of  the  poly- 
gons mentioned  in  §  460,  by  means  of  the  foregoing  formula, 
are  shown  in  the  following  table : 


REGULAR  POLYGONS  INSCRIBED  IN  A  CIRCLE  WHOSE  DIAMETER  is  UNITY 

Number  of  Sides 

Length  of  Side 

Length  of  Perimeter 

6 

.5 

3. 

12 

.25881904 

3.10583 

24 

.13052619 

3.13263 

48 

.06540313 

3.13935 

96 

.03271908 

3.14103 

192 

.01636173 

3.14145 

384 

.00818114 

3.14156 

768 

.00409060 

3.14158 

This  calculation  leads  to  the  conclusion  that 

The  circumference  of  a  circle  whose  diameter  is  one  linear 
unit  is  equal  to  3.1416  linear  units,  nearly. 

It  has  been  shown  (§  454)  that  the  perimeters  of  two  regular 
polygons  of  the  same  number  of  sides  are  to  each  other  as 
their  radii,  and  hence  also  as  the  diameters  of  their  circum- 
scribed circles.  Then  it  may  be  assumed  that  the  circumfer- 
ences C  and  C'  of  any  two  circles  have  the  same  ratio  as  their 
diameters  D  and  D'.  That  is,  C  :  C'  =  D :  D1.  Hence,  by  alter- 
nation (§  362),  C:D=C':D'.  It  appears,  then,  that  the  cir- 
cumference of  any  circle  has  the  same  ratio  to  its  diameter  as 
the  circumference  of  the  above  circle  has  to  its  diameter.  In 
other  words,  the  ratio  of  the  circumference  of  a  circle  to  the 
diameter  is  constant.  The  value  of  this  ratio  is  represented  by 


302  PLANE  GEOMETRY— BOOK  V 

the  Greek  letter  TT  (pronounced  "pi").    Hence  if  C  is  the 
circumference  of  a  circle,  D  its  diameter,  and  R  its  radius, 

C_ 

D~ 
whence  C  =  TrD,     and     C  =  2  irR. 

In  the  calculation  on  page  301,  D  =  1.  But  when  D  =  1,  the 
formula  gives  C  =  TT.  Hence  the  last  column  in  the  calcula- 
tion gives  successive  approximations  to  the  value  of  TT.  Using 
four  places  of  decimals,  we  take 

7T  =  3.1416. 

The  fraction  -^  gives  a  value  of  TT  correct  to  within  ^  of  1%. 

464.  The  length  of  an  arc  is  found  by  taking  a  proportional 
part  of  the  whole  circumference  ;  that  is, 

The  length  of  an  arc  of  n°  is x  2  wR. 

360 

EXERCISES 

1.  Draw  each  of  the  following  figures  and  prove  in  (1)  that  the 
circumference  of  the  small  circle  is  one  half  that  of  the  large  circle; 
in  (2),  that  the  sum  of  the  two  small  circumferences  equals  the  large 
circumference.  What  conclusions  may  be  drawn  from  (3),  (4),  (5)  ? 


(3)  (4)  (5) 


2.  The  length  of  a  circle  is  88  (44,  66,  a).    Find  its  radius ;  its 
diameter.    (TT  =  -2T2-.) 

3.  From  the   following    values  find   the    diameter   D  when  the 
length  of  its  circle  C  is  given,  and  find  C  when  either  R  or  D  is 
given.    C  equals  22  (40,  3  in.,  4^  ft.,  50.5  cm.,  2  km.).    D  equals 
3   (50,  2  ft.,  1£  in.,  9.1  cm.,  3  km.).    R  equals  1   (1^,  5  ft.,  4  ft., 
3£  km.,   5.1  cm.). 


CIRCUMFERENCE  OF  A  CIRCLE  303 

4.  One  of  the  famous  Sequoia  trees  in  the  Mariposa  grove  in 
California  has  a  diameter  of  32  ft.    How  large  around  is  it? 

5.  The  perimeter  of  a  circular  shaft  is  2  ft.    What  is  its  diameter  ? 

6.  The  spoke  of  a  bicycle  wheel  is  12  in.  long.    How  long  is  the 
tire  (inside  measurement)  ?    If  the  tire  is  1  in.  thick,  how  many 
times  does  the  wheel  revolve  in  traveling  5  mi.  ? 

7.  A  circular  tower  has  a  circumference  of  64  ft.    What  is  its 
diameter  ? 

8.  The  two  hands  of  a  clock  are  5  in.  and  7  in.  long  respectively. 
How  much  greater  distance  does  the  extremity  of  the  minute  hand 
travel  in  one  day  than  that  of  the  hour  hand  ? 

9.  The  radius  OA  of  the  flywheel  of  an      g-iJJLi-£JL!-5 
engine  is  8  ft.    During  each  revolution  what 

fraction  of  the  path  described  by  A  is  described  by  a  point  P,  if 
OP  is  taken  successively  equal  to  1  ft.,  2  ft.,  etc.  ? 

10.  The  diameter  of  a  circular  race  track  is  100  yd.    How  many 
laps  are  required  to  make  up  a  distance  of  10  mi.  ? 

11.  A  weight  is  to  be  lifted  by  means  of  a  wheel  and  axle.   In  the 
figure  OA  =  radius  of  wheel,  OB  =  radius  of  axle,  W  =  weight  to  be 
raised,  P  =  force    applied    at    A    necessary    to 

raise  W.    It  is  proved  in  physics  that  —  =  — -  5 

that  is,  the  force  to  be  applied  in  order  to  lift  a 
weight  is  related  to  that  weight  as  the  radius 
of  the  axle  is  to  the  radius  of  the  wheel.  If 
OB  =  -J-  0,4,  how  is  P  related  to  Wt  Find  P, 
if  W=  100  Ib.  and  OA  =  2  OB. 

12.  If  the  weight  in  Ex.  11  is  to  be  lifted  a  vertical  distance  of 
50  ft.,  and  the  diameter  of  the  axle  is  6  in.,  how  many  turns  of  the 
wheel  will  lift  the  weight? 

13.  A  rectangular  sheet  of  paper  may  be  bent  so  as  to  form  a 
circular  cylinder.    One  dimension  of  the  rectangle  then  becomes  the 
height  of  the  cylinder,  while  the  other  dimension  represents  the 
length  of  the  circle  bounding  the  base  of  the  cylinder.    How  may 
the  area  of  such  a  cylindrical  surface  be  found  ? 

14.  A  circular  tower  has  a  diameter  of  20  ft.  and  a  height  of  100  ft. 
Find  the  area  of  its  cylindrical  surface  (see  Ex.  13). 


304  PLANE  GEOMETRY— BOOK  V 

15.  In  a  certain  city  there  are  150  mi.  of  street  railways.    The 
wires  carrying  the  electric  current  have  a  diameter  of  |  in.    What 
is  the  total  surface  of  the  wires  used  for  the  trolley  service  ? 

16.  Assuming  that  in  a  certain  town  there  are  10,000  telephone 
poles,  find  the  total  surface  of  these  poles,  if  each  has  an  average 
diameter  of   9  in.  and   an   average  height  of  35  ft.   (ignoring  the 
irregular  area  of  the  tops). 

17.  In  Ex.  16,  if  these  poles  were  to  be  painted,  how  long  a  fence 
8  ft.  in  height  could  be  painted  with  the  same  amount  of  paint  ? 

18.  Assuming  the  earth  to  be  a  perfect  sphere,  how  long  is  the 
earth's  equator  if  the  radius  is  4000  mi.  ? 

In  Exs.  19-22  C  represents  the  length  of  a  circle  and  the  subscripts 
signify  different  circles. 

19.  Construct  a  circle  equal  in  length  to  the  sum  of  two  given  circles. 
Solution.    Let  x  represent  the  radius  of  the  required  circle,  while 

R  and  R'  signify  the  radii  of  the  given  circles. 

Then  2  TTX  =  2  -n-R  +  2  irR'; 

that  is,  x  =  R  +  R'- 

20.  Construct  C  =  Cl-  C2. 

21.  Construct  C  =  3  Cx  (~4  ,  | 

\   6        8 

22.  Construct  C  =  Ct  +  C2  —  C3. 

23.  If  the  radius  of  a  circle  is  10,  find  the  side  of  the  inscribed 
regular  polygons  of  4,  8,  16,  32  sides. 

24.  If  the  side  of  an  inscribed  polygon  of  2  n  sides  is  known,  the 
side  of  an  inscribed  polygon  of  n  sides  can  be  found  by  Proposition 
VIII.    Explain. 

25.  Explain  how  the  side  of  a  regular  pentagon  may  be  found 
if  that  of  the  decagon  in  the  same  circle  is  known. 

26.  Derive  formulas  for  the  sides  of  regular  inscribed  polygons  of 
8,  10,  12,  16  sides  (§  461),  if  the  radius  is  R.   Let  ss  be  the  length 
of  a  side  of  the  inscribed  polygon  of  8  sides,  etc. 


Ans.    sa  =  R  V2- 


BOOK  V 


305 


AKEA  OF  A  CIRCLE 
PRELIMINARY  DISCUSSION 

465.  The  following  exercises  will  serve  to  explain  the  topic 
to  be  considered. 

EXERCISES 

1.  Describe  on  cross-section  paper  a  circle  whose  radius  is 
two  of  the  larger  units.    Count  the  number  of  small  squares 
inclosed,  reduce  the  total  area  found  by  counting  these  squares 
to  decimals  of  a  square  whose  side  is  the  larger  unit,  and  divide 
by  the  square  of  the  radius. 

2.  Repeat  the  above  process,  using  successive  circles  with 
radii  of  three,  four,  and  five  units.    Tabulate  the  results  as 
follows : 


Radius  =  R 

Square  of  Radius 

Area=  S 

Ratio  S  :  Rz 

2 

3 

4 

5 

The  above  process  is  neither  accurate  nor  geometric.  A  com- 
putation which  is  accurate  as  well  as  geometric  may  be  made 
as  follows : 

Circumscribe  a  hexagon  about  a 
given  circle.  Bisect  each  arc  thus 
formed  and  draw  a  tangent  at  each 
bisection  point.  A  regular  circum- 
scribed polygon  of  double  the  num- 
ber of  sides  is  thus  formed.  (Why?) 
Imagine  this  process  repeated  several 
times.  The  perimeters  of  the  succes- 
sive polygons  thus  formed,  like  those  referred  to  in  §  460,  may 
be  assumed  to  approach  nearer  and  nearer  the  circumference  of 


306 


PLANE  GEOMETRY— BOOK  V 


the  circle,  and  their  areas  likewise  may  be  assumed  to  approach 
nearer  and  nearer  the  area  of  the  circle,  while  the  radius  of  the 
circle,  which  is  the  apothem  of  each  polygon,  does  not  change. 
But  it  is  known  that  the  area  of  each  polygon  is  equal  to 
half  the  product  of  its  perimeter  and  its  apothem  (§  456).  It 
follows,  then,  that  the  area  of  a  circle  is  equal  to  half  the  product 
of  its  circumference  and  its  radius. 

That  is,  S  =  ±RC. 

But  C  =  2  TrR. 


466.    The  areas  of  two  circles  are  to  each  other  as  the  squares 
of  their  radii,  or  as  the  squares  of  their  diameters. 

S       TrR2 


For 


R*_ 

nl-2 


467.  A  sector  of  a  circle  is  the  figure  formed  by  two  radii  and 
the  arc  intercepted  between  them. 

468.  The  area  of  a  sector  whose  central  angle 

O      •  n  T>2 

contains  n    is irR*. 

360 

Since  the  area  of  the  sector  bears  the  same  ratio  to 
the  area  of  the  circle  as  its  angle  bears  to  the  whole 
angular  magnitude  about  the  center,  namely,  360°. 

469.  The  area  of  a  sector  is  equal  to  one  half  the  product  of 
its  radius  and  its  arc. 

470.  A  segment  of  a  circle  is  the  figure  formed  by  an  arc 
and  the  chord  joining  its  extremities. 

The  word  "segment"  signifies  a  part  cut  off,     A^~  "^£ 

and  was  used  in  the  previous  chapters  to  mean  a 
part  of  a  line.  The  context  will  show  in  each  case 
the  sense  in  which  the  word  is  to  be  taken. 

471.  The  area  of  a  segment  of  a  circle 
(A  CB  in  the  figure)  can  be  determined  by 

finding  the  difference  between  the  area  of  the  sector  0-ACB 
and  the  area  of  the  triangle  OA  B. 


AKEA  OF  A  CIRCLE 


307 


EXERCISES 


1.  If  S  represents  the  area  of  a  circle,  which  has  the  diameter  D 
and  the  radius  R,  find  D  and  R  when  S  is  given,  and  find  S  when 
either  D  or  R  is  given,  using  the  following  values : 


S  equals 

4 

16 

36 

5  sq.  in. 

4.1  sq.  cm. 

a 

D  equals 

1 

2 

3 

^  in. 

1.2  cm. 

in  +  ft 

R  equals 

1 

2 

3 

t  in. 

3.7  cm. 

x-  y 

2.  Determine  the  effect  on  the  area  of  a  circle,  if  its  radius  is 
multiplied  by  2  ;  by  3  ;  by  #.    Does  a  similar  relation  also  hold  good 
for  diameters  ? 

3.  Find  the  area  of  a  sector  if  its  angle  and  its  radius  have  the 
following  values.    Construct  the  sector  in  each  case. 


Angle 

30° 

60° 

120° 

150° 

240° 

210° 

330° 

315° 

240° 

Radius 

2  in. 

N 

1.8  cm. 

.75  cm. 

1.15 

2.3 

.8  in. 

2.6 

.33  ft. 

4.  The  lengths  of  the  hands  of  a  watch  are  in  the  ratio  of  4  :  5. 
When  each  has  made  a  revolution,  how  do  the  areas  of  the  resulting 
circles  compare  ? 

5.  A  city  has  a  radius  of  3  mi.  How  many  acres  in  the  area  of  the  city  ? 

6.  During  an  earthquake  vibrations  were  noticed  100  mi.  from 
the  center  of  the  disturbance.   How  many  square  miles  were  exposed 
to  the  effects  of  the  shock  ? 

7.  A  man  learning  to  ride  on  a  bicycle  rides  a  distance  of  1  mi. 
the  first  day,  1^  mi.  the  next  day,  2  mi.  the  third  day,  etc.    How 
much  larger  is  the  territory  which  is  accessible  to  him  on  each  con- 
secutive day  as  compared  with  the  previous  day,  if  he  extends  his 
trips  in  this  way  for  a  week  ? 

8.  A  tree  has  a  perimeter  of  3  ft.  at  a  certain  height.    What  is 
the  area  of  its  cross  section  at  that  height? 

9.  From   a  circular  piece  of   tin  8  in.  in   diameter  the   largest 
possible  square  is  to  be  cut  (i.e.  the  inscribed  square).    How  much 
waste  tin  is  left  ? 


y  V 


308  PLANE  GEOMETRY— BOOK  V 

10.  What  is  the  answer  in  Ex.  9,  if  a  hexagonal  piece  is  to  be  cut 
out  ?  an  octagonal  piece  ? 

11.  The  volume  of  a  circular  cylinder  is  found  by  multiplying  the 
area  of  the  circular  base  by  the  length  of  the  cylinder.    What  is  the 
volume  of  a  piece  of  wire  ^  in.  in  diameter  and  20  ft.  long  ? 

12.  A  reservoir  constructed  for  irrigation  purposes  sends  out  a 
stream   of  water  through  a  pipe  3  ft.  in  diameter.    The  pipe  is 
1000  ft.  long.    How  many  times  must  it  be  filled  if  it  is  to  discharge 
10,000  acre-feet  of  water?  (An  acre-foot  of  water  is  the  amount  of 
water  required  to  cover  1  A.  to  a  depth  of  1  ft.) 

13.  An  ornamental  square  window  is  divided 
into  smaller  squares,  in  each  of  which  a  circular 
piece  of  glass  is  inserted.    Prove  that  the  com- 
bined area  of  the  circular  pieces  is  equal  to  the 
area  of  the  circle  inscribed  in  the  large  square. 

14.  If  the  diameter  of  a  water  pipe  is  doubled, 
how  many  times  as  much  water  can  the  pipe 

carry  in  a  given  time  ?   What  seems  to  be  the  relation  between  the 
increase  in  the  perimeter  of  the  pipe  and  the  increase  in  its  capacity  ? 

15.  Construct  a  circle  whose  area  shall  be  four  times  as  large  as 
that  of  a  given  circle ;  nine  times  as  large. 

In  the  following  exercises  S  signifies  the  area  of  a  circle,  while 
the  subscripts  refer  to  different  circles. 

16.  Construct  S  =  Sl  +  Sz. 

Solution.    Let  R1  and  R2  be  the  radii  of  the  given  circles,  and  let 
x  be  the  radius  of  the  required  circle. 

Then  irx2  =  VR*  +  vR* ;  that  is,  *2  =  Rf  +  R},  etc. 

17.  Construct  S  =  S1  —  S2. 

18.  Construct  S  =  5  Sr  [war2  =  5 irR*.    ,'.x*  =  5R*,  or  R:x  =  x:5 R.~] 

19.  Cgnstruct  S  =  S1  +  SZ  +  SS-  54. 

20.  In  a  circle  whose  radius  is  a,  find  the  area  of  the  smaller  seg- 
ment subtended  by  each  side  of  a  regular  inscribed  hexagon ;  of  an 
inscribed  square ;  of  an  inscribed  equilateral  triangle. 

21.  A  circular  steel  shaft  of  diameter  2  in.  is  to  be  ground  down 
so  that  its  diameter  becomes  1£  in.    What  fraction  of  its  volume  is 
to  be  removed? 


BOOK  V  309 

MENSUEATION  OF  THE  CIECLE.    FOEMAL 
DEMONSTEATION 

VARIABLES  AND  LIMITS 

472.  In  counting  the  number  of  books  on  a  bookshelf  one  con- 
stantly separates  the  total  number  of  books  into  two  groups,  those 
counted  and  those  not  counted.  The  number  of  books  which  have 
actually  been  counted  is  continually  increasing,  while  the  number 
of  books  that  are  still  to  be  counted  is  continually  decreasing. 
The  total  number  of  books,  however,  remains  the  same.    This 
illustration  serves  to  make  clearer  the  following  definitions : 

473.  A  number  which  has  different  values  during  the  same 
discussion  is  called  a  variable. 

474.  A  number  which  retains  the  same  value  throughout  a 
discussion  is  called  a  constant. 

12345678 

Let  OA  represent  a  rod  8  in.  long.    Imagine       0~P~°^  °~A 

the  point  P  to  move  from  0  to  A.    Then  the 

numbers  expressing  the  lengths  of  the  segments  OP  and  PA  are  vari- 
ables, while  the  total  length  OA  is  a  constant. 

475.  Cut  a  rectangular  piece  of  paper  into  two  equal  parts. 
Lay  one  of  these  parts  aside  and  bisect  the  other.    Lay  one  of 
these   last   two  pieces  aside  and   bisect  the  one  remaining. 
Eepeat  this  process  a  number  of  times.     If  the  area  of  the 
original  piece  is  one  square  unit,  then  the  areas  of  the  pieces 
laid  aside  are  1,  J,  £,  y1^,  etc.,  of  a  square  unit.    At  any  stage 
in  the  process  the  total  area  of  all  these  pieces  will  differ  from 
the  area  of  the  original  piece,  namely,  one  square  unit,  by  the 
area  of  the  piece  about  to  be  bisected.    The  area  of  the  piece 
about  to  be  bisected  is  successively  £,£,£,  yV?  e^c-    By  repeated 
bisection,  the  area  of  this  piece  ivill  become  and  remain  less  than 
any  assigned  positive  number,  however  small. 

Thus  it  appears  that  the  area  of  the  piece  bisected  is  a 
variable  which  approaches  0,  while  the  total  area  of  the  other 
pieces  is  a  variable  which  approaches  1. 


310  PLANE  GEOMETRY— BOOK  V 

This  illustration  will  serve  to  introduce  the  definition  of  the 
next  section. 

476.  Limit  of  a  Variable.    When  the  successive  values  of  a 
variable  approach  a  certain  constant  number  so  that  the  differ- 
ence between  the  constant  and  the  variable  becomes  and  remains 
less  than  any  assigned  positive  number,  however  small,  then  the 
constant  is  called  the  limit  of  the  variable. 

477.  The  statement  "  x  approaches  the  limit  a,"  where  x  is 
a  variable  and  a  is  a  constant,  is  sometimes  written 

x  =  a, 
the  symbol  ==  meaning  "  approaches  the  limit." 

APPLICATION  OF  THE  THEORY  OF  LIMITS  TO  THE 

DETERMINATION  OF  THE  LENGTH  AND  THE 

AREA  OF  A  CIRCLE 

478.  About  any  given  circle  circumscribe  a  square.   By  join- 
ing the  points  of  tangency  an  inscribed  square  is  obtained. 
(Why  ?)     The  perimeter  of  the  inscribed 

square  is  less  than  that  of  the  circumscribed 
square.    (Why  ?) 

Suppose  now  that  the  arcs  subtended 
by  the  sides  of  the  inscribed  square  are 
bisected,  and  that  each  point  of  bisection 
is  joined  to  the  adjoining  vertices  of  the 
inscribed  square.  A  regular  inscribed  poly- 
gon of  eight  sides  is  thus  constructed,  the  perimeter  of  which, 
as  can  be  shown  by  the  methods  of  inequalities  (Exs.  6  and  9, 
p.  294),  is  greater  than  that  of  the  inscribed  square,  but  less 
than  that  of  the  circumscribed  square.  If  this  process  of 
bisecting  arcs  is  repeated  several  times,  a  series  of  regular 
inscribed  polygons  is  constructed.  The  perimeters  of  these 
polygons  increase  step  by  step,  but  they  always  remain  less 
than  the  perimeter  of  the  circumscribed  square. 


MENSURATION  OF  THE  CIRCLE  311 

In  like  manner  the  areas  of  the  successive  polygons  increase 
but  always  remain  less  than  the  area  of  the  circumscribed 
square. 

479.  Suppose  now  that  tangents  are  drawn  at  the  middle 
points  of  the  arcs  subtended  by  the  sides  of  the  inscribed 

square,  to  meet  the  sides  of  the  circum- 

scribed  square.  Then  a  circumscribed  poly- 
gon of  eight  sides  results.  Suppose,  further, 

that  this  process  of  drawing  tangents  is 

continued,   the   number   of   sides    of    the 

circumscribed  polygon  being  doubled  again 

and  again.    It  can  be  shown  (Exs.  7  and  8, 

p.  294)  that  the  perimeters  and  areas  of 

the   successive   circumscribed   polygons   decrease  but  always 

remain  greater  than  the  perimeter  and  the  area  respectively 

of  the  inscribed  square. 

480.  It  is  apparent  that  the  above  considerations  are  inde- 
pendent of  the  number  of  sides  of  the  regular  circumscribed 
polygon  selected  at  the  outset  (Ex.  10,  p.  294). 

481.  In  the  first  case  the  perimeter  and  the  area  of  the 
regular  inscribed  polygon  are  increasing  variables,  which  are 
always  less  than  the  constant  perimeter  and  the  constant  area 
respectively  of  a  definite  regular  circumscribed  polygon ;  while 
in  the  second  case  the  perimeter  and  the  area  of  the  regu- 
lar circumscribed  polygon  are  decreasing  variables,  which  are 
always  greater  than  the  constant  perimeter  and  the  constant 
area  respectively  of  a  definite  regular  inscribed  polygon. 

Dropping  the  geometrical  language,  we  have,  in  the  first 
instance,  two  variables,  each  of  which  constantly  increases 
but  remains  less  than  a  given  number;  and  in  the  second 
instance,  two  variables,  each  of  which  constantly  decreases 
but  remains  greater  than  a  given  number.  The  connection 
with  the  Theory  of  Limits  is  afforded  now  by  the  following 
assumption : 


312  PLANE  GEOMETRY— BOOK  V 

482.  Axiom.    Existence  of  a  Limit.    If  a  variable  constantly 
increases  but  always  remains  less  than  a  given  constant,  then  the 
variable  approaches  a  limit.  This  limit  is  either  less  than  or  equal 
to  the  given  constant. 

Thus  if    the    point    P     A~      + + + G 

assumes  different  positions 

on  the  line  AC  so  that  AP  always  increases  but  remains  less  than  AB, 
then  the  variable  length  AP  approaches  a  limiting  value  AL,  which  is 
less  than  or  equal  to  AB. 

483.  It  follows  that  if  PC  in  the  above  figure  constantly 
decreases  but  always  remains  greater  than  BC,  then  the  vari- 
able length  PC  approaches  a  limit  LC,  which  is  greater  than 
or  equal  to  EC.   That  is,  as  a  consequence  of  the  above  axiom, 

If  a  variable  constantly  decreases  but  always  remains  greater 
than  a  given  constant,  then  the  variable  approaches  a  limit  which 
is  greater  than  or  equal  to  the  given  constant. 

484.  In  accordance  with  the  conclusion  of  §  481  and  the 
axiom  of  §  482,  the  perimeter  of  a  regular  polygon  inscribed 
in  a  given  circle  will  approach  a  definite  limit  when  the  num- 
ber of  its  sides  is  doubled  an  indefinite  number  of  times ;  and 
the  area  of  this  polygon  also  will  approach  a  definite  limit. 

Similarly,  by  the  preceding  section,  the  perimeter  of  a  regular 
polygon  circumscribed  about  a  given  circle  will  approach  a  defi- 
nite limit  when  the  number  of  its  sides  is  doubled  an  indefinite 
number  of  times,  and  the  area  also  will  approach  a  definite  limit. 

485.  It  can  now  be  shown  that  the  perimeter  of  the  inscribed 
polygon  and  the  perimeter  of  the  similar  circumscribed  polygon 
will  approach  the  same  limit.   This  is  done  by  proving  that  the 
ratio  of  the  perimeters  of  the  two  polygons  approaches  unity 
as  a  limit. 

Proof.  In  the  figure  let  A  B  and  A  'B'  be  homol- 
ogous sides  of  two  similar  regular  polygons  in- 
scribed in  and  circumscribed  about  the  same  circle. 
Denote  AB  by  «,  and  OC'  by  R,  and  the  perimeters 
of  the  two  polygons  by  JP  and^?'  respectively. 


MENSURATION  OF  THE  CIRCLE  313 


Then       P  -  -.         Why? 

p'  ~  OC'  ~  R  R 

As  the  number  of  sides  is  increased,  the  side  a  continually 
decreases  and  approaches  zero  as  a  limit,  the  radius  R  remain- 
ing constant.  The  ratio  —  increases  but  remains  less  than  unity. 
Hence  this  ratio  approaches  a  limit  (§  482),  and  the 

limit  ofj,=  -  — =  ^  =  1. 

That  is,  p  and_2/  have  the  same  limit. 

486.  In  like  manner  the  areas  of  the  two  polygons  may  be 
shown  to  approach  the  same  limit. 

The  foregoing  discussion  justifies  the  following  definitions : 

487.  The  circumference  of  a  circle  is  the  common  limit  ap- 
proached by  the   perimeters  of  a  regular  inscribed    polygon 
and  the  similar  regular  circumscribed  polygon,  as  the  number 
of  sides  is  continually  increased. 

488.  The  area  of  a  circle  is  the  common  limit  approached  by 
the  areas  of  these  polygons. 

REMARK.  There  remains,  however,  the  following  difficulty.  Begin- 
ning with  the  inscribed  square,  by  bisecting  arcs  and  proceeding  as  in 
§  478,  a  series  of  regular  inscribed  polygons  of  4,  8, 16,  32,  etc.  sides  may 
be  constructed  and  the  perimeters  will,  by  §  482,  approach  a  definite  limit. 
Again,  beginning  with  the  regular  inscribed  hexagon,  the  same  construc- 
tion leads  to  a  series  of  regular  inscribed  polygons  of  6,  12,  24,  48,  etc. 
sides,  and  their  perimeters  also  will  approach  a  definite  limit.  Is  this  limit 
the  same  for  each  of  the  two  series  of  polygons?  The  definition  of  §487 
assumes  that  these  limits  are  the  same.  The  proof  would  not  be  under- 
stood at  this  time,  and  it  is  sufficient  here  to  point  out  that  the  assumption 
indicated  lurks  in  the  definitions  of  §§  487  and  488. 

489.  The   following   conclusion  is   used  in  the   proofs  of 
Propositions  IX  and  X :  If  the  variable  x  approaches  the  limit 


x 


t 


a,  then  ^  x  will  approach  the  limit  1  a,  and,  in  general,  '—  will 

approach  the  limit  —  >  c  being  any  constant  (not  zero). 
c 

The  proof  follows  immediately  from  the  definition  of  §  476. 


314  PLANE  GEOMETRY— BOOK  V 

PROPOSITION  IX.    THEOREM 
490.   Two  circumferences  have  the  same  ratio  as  their  radii. 


Given  two  circles  with  circumferences  C  and  Cf,  and  with  radii 
R  and  R1  respectively. 

To  prove  that  C  :  C'  =  R  :  R'. 

Proof.  1.  Inscribe  in  each  circle  a  regular  polygon  of  n  sides, 
and  let^>  andj>'  be  their  perimeters. 

2.  Then  p\p*  =  R\n\  Why? 

R  =  R<' 

3.  Let  n  increase  indefinitely. 

Then  |~i'  §§487,489 

a     R 

p<   .    C' 
-R-  =  R>- 

7}  79 

But  ^-  and  =^>  being  always  equal  variables,  are  really  one 
A  A 

and  the  same  variable.    Hence  their  limits  are  equal. 

C       C' 
:t-5     Jf!1 

or  C:C'  =  R:R'. 

491.  COROLLARY  1.   Two  circumferences  have  the  same  ratio 
as  their  diameters. 

492.  COROLLARY  2.   The  ratio  of  a  circumference  to  its  di- 
ameter is  constant. 

This  constant  ratio  is  denoted  by  the  letter  TT  (§  463). 


MENSURATION  OF  THE  CIRCLE  315 

PROPOSITION  X.    THEOREM 

493.   The  area  of  a  circle  is  equal  to  half  the  product  of  its 
radius  and  its  circumference. 


Given  the  circle  0,  of  which  the  area  is  S,  the  radius  1?,  and  the 
circumference  C. 

To  prove  that  S  =  %  C  x  R. 

Proof.  1.  Circumscribe  a  regular  polygon  of  n  sides  about 
the  circle,  and  let  p  be  its  perimeter  and  A  its  area.  The 
apothem  is  equal  to  the  radius  of  the  circle. 

2.  Then  A  =  \p  x  R.  §  456 

3.  Let  the  number  of  sides  of  this  regular  circumscribed 
polygon  increase  indefinitely. 

Then  A  =  S,  §  488 

and  %px  R±iCx  R.  §§487,489 

But  A  and  \p  x  R,  being  always  equal  variables,  are  really 
one  and  the  same  variable.  Hence  their  limits  are  equal. 

4.  .'.S=±CxR. 

494.  COROLLARY  1.   The  area  of  a  circle  is  equal  to  TT  times 
the  square  of  the  radius. 

495.  COROLLARY  2.   The  areas  of  two  circles  are  to  each  other 
as  the  squares  of  their  radii  or  of  their  diameters. 

Discussion.  Why  is  the  circumscribed  polygon  used  in  the  above 
theorem,  while  the  inscribed  polygon  is  used  in  Proposition  IX  ? 


316  PLANE  GEOMETRY— BOOK  V 

496.  Historical  Note.  The  development  of  geometry  was  aided  very 
decidedly  by  the  appearance  of  problems  whose  solution  seemed  to  defy 
the  powers  of  the  best  mathematicians.  Among  these  difficult  problems 
none  has  exerted  a  greater  influence,  or  has  attracted  greater  attention 
throughout  the  ages,  than  the  famous  problem  of  "the  squaring  of  the 
circle.1'  In  its  original  form  the  problem  meant  the  construction  of  a 
square  which  should  have  an  area  equal  to  that  of  a  given  circle.  As 
will  now  be  readily  understood,  this  question  involves  the  determination 
of  TT,  or  the  construction  of  a  line  equal  to  the  circumference  of  the  given 
circle.  The  history  of  this  problem  extends  over  a  period  of  four  thousand 
years,  and  may  be  divided  into  three  periods : 

1.  From  the  earliest  times  to  the  17th  century  A.D.  During  this  time  the 
value  of  TT  was  computed  by  means  of  polygons,  as  in  Proposition  VIII. 

2.  From  the  beginning  of  modern  mathematics  (calculus,  etc.]  to  1766. 
During  this  period  the  value  of  TT  was  computed  more  accurately  by 
algebraic  methods. 

3.  The  modern  period.   In  1766  Lambert  proved  that  TT  is  irrational. 
This  prepared  the  way  for  the  modern  discovery  that  TT  cannot  be  con- 
structed with  ruler  and  compasses.   The  proof  was  given  by  Lindemann 
in  1882.    Hence  it  is  impossible  to  construct  geometrically  a  line  equal  to  the 
circumference  of  a  given  circle,  or  a  square  having  an  area  equal  to  that  of  a 
given  circle. 

The  earliest  reference  to  the  number  IT  is  found  in  the  Rhind  Papyrus 
(British  Museum),  written  by  Ahmes,  an  Egyptian  scribe,  about  1700  B.C. 
His  rule  for  finding  the  area  of  a  circle  consists  in  squaring  eight  ninths 
of  the  diameter,  which  makes  TT  =  3.1604. 

The  Bible  mentions  the  number  TT  in  two  places  :  1  Kings  vii,  23  ; 
and  2  Chronicles  iv,  2,  giving  TT  =  3  (probably  a  Babylonian  method). 

Archimedes  (born  287  B.C.),  the  greatest  mathematician  of  antiquity, 
by  a  method  similar  to  that  of  Proposition  VIII,  proved  that  the  value  of 
Trlies  between  3^  and  3^j,  or,  in  decimals,  between  3.1428  and  3.1408. 

Ptolemy,  a  great  astronomer  (150  A.D. ),  found  that  TT  =  3TyT  =  3.14166. 

The  Hindus  used  TT  =  VlO  =  3.1623  ;  also,  TT  =  f  f  f  J  =  3.1416. 

Metius  of  Holland  (1625  A.D.)  found  TT  =  f  f  f  =  3.1415929,  and  this 
decimal  is  correct  through  the  sixth  place. 

Ludolph  van  Ceulen  (Leyden,  1610)  carried  the  value  of  TT  to  thirty- 
five  decimal  places. 

By  improved  methods  of  calculation  the  value  of  TT  has  been  carried 
in  recent  years  by  Shanks  to  707  decimal  places.  The  symbol  TT,  in  the 
present  sense,  was  used  for  the  first  time  by  William  Jones,  in  1706.  It 
can  be  shown  that  TT  is  neither  a  rational  fraction  nor  a  surd.  It  cannot 
be  expressed  exactly  by  an  algebraic  number.  It  is  therefore  called  a 
transcendental  number. 

The  correct  value  of  TT  to  ten  places  of  decimals  is  3.1415926535. 


MENSURATION  OF  THE  CIRCLE  317 

EXERCISES 

1.  In  a  certain  city  whose  diameter  is  about  10  km.,  it  is  desired 
to  construct  a  circular  boulevard  around  the  outskirts.    Estimate  the 
errors  in  the  length  of  the  inner  curb  which  would  arise  from  using 
the  successive  approximations  to  the  value  of  TT  in  §  463. 

2.  Plot  a  graph  showing  successive  approximations  of  TT  (§  463). 
Suggestion.  On  the  horizontal  axis  lay  off  distances  representing  the 

number  of  sides  in  the  polygons  used,  and  on  the  vertical  axis  lay  off 
the  corresponding  values  of  TT.   Use  a  large   scale  on  the  vertical  axis. 

3.  Many  attempts  have  been  made  to  construct  a  line  equal  to  the 
length  of  a  circle.  The  following  approximate  construction  is  one  of 
the  simplest.    It  is  due  to  Kochansky  (1685). 

At  the  extremity  A  of  the  diameter  AB  of 
a  given  circle  of  radius  R  draw  a  tangent  CD, 
making  Z  CO  A  =30°  and  CD  -  3  /?.  Prove 
that  BD  =  R  Vl3^-2V3  -  3.1415  A' ;  that 
is,  BD  =  |  the  circumference,  very  nearly. 

4.  Archimedes  (250  B.C.)  stated  the  prop- 
osition that  the  area  of  a  circle  is  equal  to  that  of  a  triangle  whose 
base  is  the  length  of  the  circle,  and  whose  altitude  is  the  radius. 
Explain.    Construct  this  triangle  approximately  by  using  Ex.  3. 

5.  Since  it  is  possible  to  transform  any  triangle  into  a  square, 
construct  a  square  approximately  equal  in  area  to  that  of  a  circle  by 
using  the  method  of  Ex.  3. 

6.  A  very  old  Egyptian  manuscript,  written  about  1700  B.C.,  con- 
tains this  rule  for  finding  the  area  of  a  circle ;  that  is,  for  "  squaring 
a  circle."    From  the  diameter  of  a  circle  subtract  one  ninth  of  the 
diameter,  and  square  the  remainder.    To  what  value  of  TT  does  this 
construction  correspond  ? 

7.  Hippocrates  (430  B.C.),   a  great  Greek  geome- 
trician, tried  to  "  square  the  circle."    The  following 
theorem  illustrates  his  method  of  approaching  the 
problem :   If  on  the  sides  of  an  inscribed  square  as 
diameters  semicircles  are  described,  the  area  of  the 

four  crescents  lying  without  the  circle  equals  the  area  of  the  in- 
scribed square.    Give  proof. 


318  PLANE  GEOMETRY— BOOK  V 

REVIEW   EXERCISES 

1.  What  are  the  important  topics  of  Book  V  ? 

2.  What  use  is  made  of  regular  polygons  in  Book  Y  ? 

3.  Give  a  brief  description  of  the  process  by  which  the  length 
and  the  area  of  a  circle  are  determined  geometrically. 

4.  Define  TT.    Prove  that  it  is  a  constant. 

5.  Give  two  formulas  for  the  length  of  a  circle. 

6.  Give  three  formulas  for  the  area  of  a  circle. 

7.  How  is  the  area  of  a  sector  found  ? 

8.  How  is  the  area  of  a  segment  found? 

9.  If  the  radius  of  a  circle  is  multiplied  by  2  (3,  4,  n),  what  is 
the  effect  on  the  length  of  the  circle  ?  on  the  area  ? 

10.  The  areas  of  two  regular  octagons  are  as  1  : 16,  and  the  sum 
of  their  perimeters  is  25.    How  long  is  a  side  of  each  ? 

11.  A  side  of  a  regular  hexagon  is  12.    Find  the  radius  of  the 
inscribed  circle. 

12.  A  circle  and  a  square  have  equal  areas.   Which  has  the  greater 
perimeter  ? 

13.  A  circle  and  a  square  have  equal  perimeters.    Which  has  the 
greater  area  ? 

14.  The  area  of  a  circle  is  to  be  divided  into  3  (4,  5,  n)  equal  parts 
by  means  of  concentric  circles.    If  the  radius  of  the  given  circle  is 
100,  what  are  the  radii  of  the  concentric  circles  ? 

15.  The  arch  of  a  bridge  has  the  form  of  a  circular  arc.    Its  span 
(chord)  is  280  ft.,  and  the  greatest  height  of  the  circular  arc  above  its 
chord  is  80  ft.   Find  the  radius  of  the  circle  of  which  the  arc  is  a  part. 

16.  The  radius  of  a  circle  is  10.   Find  the  area  lying  between  two 
parallel  sides  of  the  inscribed  regular  hexagon. 

17.  The  apothem  of  a  regular  hexagon  is  4.    Find  the  area  of 
the  hexagon. 

18.  Find  the  number  of    degrees   in  the    central 
angle   of    a   sector  if    its  perimeter  is  equal  to  the 
circumference  of  the  circle  of  which  it  forms  a  part. 

19.  Show  by  the   figure  that  the  value   of    C :  D 

lies  between  3  and  4.    (Use  the  perimeters  of  the  polygons.) 


MENSURATION  OF  THE  CIRCLE 


319 


MISCELLANEOUS  EXERCISES 

COMPOSITE  FIGURES 

TREFOILS 

The  equilateral  triangle  may  be  used  as  the  foundation  of  numer- 
ous ornamental  designs,  which  are  called  trefoils. 

These  figures  are  often  seen  in  decorative  patterns,  and  are  fre- 
quently introduced  in  the  construction  of  church  windows. 

1.  Construct  an  equilateral  triangle.  With  each  vertex  as  a 
center,  and  with  one  half  of  a  side  as  a  radius,  describe  arcs  as  in- 
dicated in  Figs.  1  and  2.  Let  2  a  represent  the  length  of  a  side  of  the 
equilateral  triangle.  Find  the  perimeter  and  the  area  of  the  figure 
bounded  by  the  arcs. 


FIG.  1 


FIG.  2 


FIG.  3 


FIG.  4 


2.  Modify  the  preceding  exercise  by  using  the  mid-points  of  the 
sides  as  centers,  as  indicated  in  Figs.  3  and  4. 

3.  Inscribe  an  equilateral  triangle  in  a  circle 
of  radius  2  a.   Using  the  mid-point  of  each  radius 
of  the  triangle  as  a  center,  and  a  as  a  radius, 
describe  circles.    A  symmetric  pattern  will  result. 
Find  the  perimeter  and  the  area  of  the  trefoil 
and  of  the  shaded  part  of  the  figure. 

4.  The  perimeter  of  a  certain  church  window  is  made  up  of  three 
equal  semicircles  the  centers  of  which  form  the  vertices  of  an  equi- 
lateral triangle  which  has  sides  3jft.  long.    Find  the  area  of  the 
window  and  the  length  of  its  perimeter.    (Harvard  College  Entrance 
Examination  paper.) 

5.  If  the  area  of  the  trefoil  in  Fig.  3  above  is  50  sq.  ft.,  how  long 
is  one  side  of  the  equilateral  triangle  ?    (Take  TT  =  272 .) 

6.  Assuming  that  the  area  of  the  trefoil  in  each  of  Figs.  1,  2,  and  4 
above  is  50  sq.  ft.,  find  in  each  case  one  side  of  the  equilateral  triangle. 


320 


PLANE  GEOMETRY— BOOK  V 


QUATREFOILS 

A  square  may  be  used  as  the  foundation  of  ornamental  figures, 
which  are  usually  called  quatrefoils. 

7.  Construct  a  square.  The  length  of  a  side  is  2  a.  In  Figs.  1 
and  2  use  the  vertices  as  centers  and  one  half  of  a  side  as  a  radius. 


FIG.  1 


FIG.  2 


FIG.  3 


FIG.  4 


In  Figs.  3  and  4  use  the  mid-point  of  each  side  as  a  center.  In  each  case 
find  the  perimeter  and  the  area  of  the  figure  bounded  by  the  arcs. 
(The  construction  lines  in  Fig.  3 
show  how  the  area  of  one  lobe  of 
the  quatrefoil  may  be  found.) 

8.  The    adjoining    quatrefoil 
arises    from    a    combination    of 
Figs.  2   and  3    of  the  preceding 

exercise.    Find  the  area  and  the  perimeter  of  the  figure  bounded  by 
the  arcs.     (The  construction  lines  indicate  an 
equilateral    triangle,   from    which    the    relation 
of  the  arcs  can  be  inferred.) 

9.  In  the  adjoining  figure  find  the  perimeter 
and   the    area    of   ABCD---K,   if   the    side   of 
the  square  is  2  a  and  the  radius  of  the  circle 
is  ^  a. 

10.  The  adjoining  cross-shaped  figures  are  obtained  by  using  the 
vertices  of  the  square  as  centers 
and  one  half  of  a  diagonal  as 
a  radius.  Find  the  perimeter  and 
the  area  of  each  figure  if  the 
diagonal  equals  2  a. 

(The  second  figure  was  given  on 
page  230  for  the  purpose  of  determining  the  area  of  a  regular  octagon.) 


MENSURATION  OF  THE  CIRCLE 


321 


11.  In  the  adjoining  double  quatrefoil  the  triangle  ABC  is  equi- 
lateral.   Find  the  area  and  the 

perimeter  of  the  pattern,  mak- 
ing no  allowance  for  the  over- 
lapping of  the  two  strips.  The 
arcs  of  each  strip  are  concentric, 
their  centers  being  the  vertices 
of  the  square.  (Observe  that  the 
altitude  of  &ABC  is  known.  Then  the  length  of 
AB  can  be  found.) 

12.  On  a  network  of  equal  squares  of  side  a  con- 
struct the  vaselike  figure   shown  in  the   diagram. 
The  center  of  each  arc  is  at  the  center  of  a  square. 
Find  the  perimeter  and  the  area  of  the  figure. 

13.  The  centers  of  the  four  arcs  in  the  adjoin- 
ing figure  are  the  vertices  of  the  square.    Prove  that 
A  BCD  is  a  square,  and  find  its  area,  each  side  of 
the  larger  square  being  2  «.    Find  also  the  area  of 
the  quatrefoil  whose  vertices  are  A,  B,  C,  and  D. 

Suggestion.    Observe    that  AD  subtends   an  arc 
of  30°. 

MULTIFOILS 

Any  regular  polygon  may  be  used  for  the  construction  of  figures 
similar  to  those  suggested  for  the  square. 

14.  A  rose  window  of  six  lobes  is  to  be  placed 
in  a  circular  opening  14  ft.  in  diameter.    It  is  to 
have  the  form  shown  in  the  diagram.    Determine 
the  radius  of  the  interior  hexagon ;  the  perimeter 
of  the  hexafoil ;  the  area  of  the  .hexafoil. 

(How  does  a  radius  of  one  of  the  smallest 
circles  compare  in  length  with  a  radius  of  the 
largest  circle  ?) 

15.  About  the  vertices  of  a  regular  inscribed 
hexagon  as  centers,  and  with  radii  equal  to  the 
radius  of  the  given  circle,  describe   arcs   within 
the  circle.   If  the  radius  of  the  circle  is  a,  find  the 
perimeter  and  the  area  of  the  resulting  hexafoil. 


322 


PLANE  GEOMETRY— BOOK  V 


ARCHES 


Arches  are  of  great  importance  in  architecture.  The  most  common 
forms  are  semicircular,  segmental,  and  pointed. 

In  the  diagrams  AB  is  the  width  or  span  of  the  arch,  while  CD 
is  its  height.  Owing  to  the  symmetry  of  these  arches,  numerical 


computations  involving  these  forms  are  greatly  simplified.    In  the  fol- 
lowing problems,  unless  the  context  suggests  a  different  meaning,  let 

s  =  span  =  AB,     h  =  height  =  CD, 

p  =  perimeter  of  the  arch  =  length  of  arc  A  D  +  arc  BD, 
A  =  area  of  the  arch  =  area  inclosed  by  p  and  s. 

16.  Draw  a  semi- 
circular  arch.    Let 
s  =  a.    Find  h,  p,  A . 

17.  The  Arch  of 
Triumph    in    Paris 
is   162  ft.    in   total 
height,    147  ft.    in 
width,  and  73  ft.  in 
depth.    The  crown 
of     its    vast    arch, 
which  is  semicircu- 
lar,   is   96  ft.   from 
the     ground.     The 
width   of    the  arch 
is  48  ft.    Find  the 
perimeter     of     the 
arch ;    the   area   of 
the  interior  passage 

of  the  arch;  the  entire  exterior  surface  of  the  structure,  disregard- 
ing cornices  and  other  projections. 


MENSURATION  OF  THE  CIRCLE 


323 


18.  In  the  adjoining  design  of  a  window  let  AB  =  4  a.    Find  the 
length  of  each  arc  and  the  area  of  each  part 

of  the  figure. 

19.  If   AB  =  4  a  in   the   following  design, 
find  the  radius  FD  of  the  upper  circle,  and 
then  determine  the   area  lying  between  the 
circles. 

Analysis.  Let    FD  —  x. 


Then 
But 


(a 


CF  =  2  a  -  x. 
EF*  =  CE2  + 


2a 


A     E 


A    E 


F  B 
circles. 


10 


20.  In   the    adjoining   window    design   let 
AB  —  6  a.     0    and    0'   are    the    vertices    of 
equilateral  triangles  constructed  on  CE  and 
CF  respectively.    Find  the  area  of  circles  0 

and   0',  and  the  total   area  of  the  crescents  between  the 

21.  A  segmental  arch  over  a  door  subtends  a  cen- 
tral angle  of  60°.    If  the  door  is  10  ft.  high  and  4  ft. 
wide,  find  h,  p,  and  A . 

22.  The  equilateral  Gothic  arch  has  already  been  de- 
fined (Ex.  13,  p.  190).    Find  the  area  of  an  equilateral 
Gothic  arch  if  s  =  12. 

23.  The  horseshoe  arch  is  used  extensively  in  Moorish 
architecture.    With  AC  1=  -\  as  a  radius  and  A 

as  a  center,  construct  a  quadrant  CF.  Trisect 
arc  CF  at  D  and  E.  Draw  AD  and  produce  it 
to  meet  the  perpendicular  bisector  of  AB  at  //. 
From  H  as  a  center,  with  radius  HA,  draw  a  cir- 
cle intersecting  CH  produced  at  0.  Then  0  is 
the  center  of  the  arch  (radius  OA).  If  s  =  2  a, 
find  h,  p,  and  A. 

Suggestion.  In  the  rt.  A  A  CH,  Z  A  =  30°  and  A  C  =  a.    Find  CH 
and  prove  OA  =  HA. 


324 


PLANE  GEOMETRY— BOOK  V 


/H 


--* -^?— IB 


24.  The  seymental  pointed  arch  has  its  centers  below  the  span.    In 
the  figure  CM  =  ^  OA.    C  and  D  are  the  centers  of 

the  arcs  BE  and  AE  respectively.    If  s  =  4  a,  and 
OM  =  by  find  ME. 

Suggestion.  CE  =  CB.    Observe  that  CB  is  the 
hypotenuse  of  a  right  triangle  with  legs  I  and  3  a. 

25.  A  four-centered   arch    may    be    constructed 
as  follows :    Divide  the   span    into   four  equal 
parts.    On   CD  construct  a  rectangle,  making        , 
CF  =  \AB.     Draw    the    lines    FH   and    EK. 

C  and  D  are  the  centers  of  the  small  arcs, 
and  E  and  F  of  the  larger  arcs.  If  s  =  4  a, 
find  EK. 

26.  The    Brescia    arch    (Turkish)     is    con- 
structed by  dividing  s  into  8  equal  parts.    On 
AC  and  BD,  each  containing  three  of  these 
parts,    equilateral    triangles    are    constructed. 
The  arcs  AF  and  EB  have  their  centers  at  C 
and  D  respectively.  At  F  and  E  draw  tangents 
to  these  arcs  meeting  in  H.    If  s  =  8  a,  find  h,p, 
and  A.    (Draw  FE.) 

27.  The    figure    represents    a    Persian  arch. 
Triangles  ABC  and  DEF  are  congruent  and 
equilateral.    The  centers  of  the  upper  arcs  MC 
and  NC  are   respectively  D  and  7%  while  the 
lower  arcs  are  drawn  with  the  center  E.    Prove 
that  the  area  of  the  arch  equals  the  area  of 
triangle  ABC. 

28.  In    the    adjoining    modification   of    the 
Persian  arch  the  four  arcs  are  equal  quadrants. 
If  s  —  4  a,  prove  that  the  area  of  the  arch  is 
4  a2. 

29.  Draw  the   outline   of  window  tracery  in  the 
figure,    using   the    given    dimensions.     The    arch   is 
equilateral,   and  all    the   arcs    are  of  equal    radius. 
Find  the  height  of  each  of  the  pointed  arches  within 
the  equilateral  arch. 


D 


MENSURATION  OF  THE  CIRCLE 


325 


30.  In   the   adjoining    diagram,  representing  a   Gothic   window, 
AB  =  EC  =  CD  =  DE.    The  arches  are  all 

equilateral.  The  point  0  is  the  intersection 
of  arcs  having  the  radii  A  D  and  EB  (Ex.  14, 
p.  190).  The  construction  of  the  trefoil  is  ex- 
plained in  Ex.  3.  If  AB  =  a,  find  the  area  of 
each  part  of  the  figure,  neglecting  the  tracery. 

31.  The    drop-pointed   arch    is    formed    by 
two  arcs  whose  radii  are  less  than  the  span. 
In  the  figure,  AB  is  trisected  at  C  and  D. 
The  arcs  are  constructed  with  C  and  D  as  centers, 
and  radii   CB  and  DA  respectively.     If  s  =  6  «, 
find  h.     Prove  that  h  =  .645  x  s  approximately. 

32.  The  Early  English  or  lancet  arch  is  formed 
by  two   arcs   whose    radii   are    greater  than  the 
span.    In  the  figure  the  span  AB  is  bisected 
at  C.  Construct  the  squares  CE  and  CF.  Make 
CH  =  CE,  and  CK  =  CF.    The  arcs  are  con- 
structed with  H  and  K  as  centers,  and  with 
radii  II B  and  KA  respectively.    If  AB  =  2  a, 

find  h.    Prove  that  h  =  .979  x  s  approximately.       HA        C        B  K 


MOLDINGS  AND  SCROLLS 

The  construction  of  moldings  and  scrolls  is  based  very  largely  on 
the  principles  of  circles  that  are  tangent  internally  or 
externally  (see  Book  II,  Proposition  XIV).    These 
designs  are  of  importance  in  a  number  of  industries 
and  trades. 

33.  Draw  the  scotia  molding  shown  in  the  margin. 
The  curve  is  made  up  of  two  quadrants  of  1  in.  and 
^  in.  radius  respectively.    How  long  is  the  curve  ? 

34.  Draw  the  cyma  recta  molding  shown  in  the 
margin,  using  the  given  dimensions.    The  curve  is 
composed  of  two  quadrants  of  equal  radii,  tangent 
to  each  other  and  to  the  lines  AB  and  C-D  respec- 
tively.   How  long  is  the  curve  ? 


326 


PLANE  GEOMETRY— BOOK  V 


35.  Draw  the   accompanying  diagram  of  window  tracery,  using 
the    given    dimensions.     The    arch    is    equilateral. 

Explain   how   the    radius   of  the   interior  circle  is 
found. 

36.  Copy  the  annexed  scroll   (figure   below).    It 
consists  of  two  spirals  and  a  connecting  arc.    The 
spirals  have  two  centers,  used  in  succession  (Ex.  9, 
p.  163).  The  connecting  arc  is  drawn  with  its  center 
at  0,  the  vertex  of  an  equilateral   tri- 
angle BOD.   If  A B  =  5  mm.,  and  BD  = 

7  cm.,  find  the  length  of  the  scroll. 

37.  Modify  the  figure  in  the  previous 
problem  by  connecting  the  two  spirals 
by  a  straight  line  tangent  to  each  spiral. 

38.  A    balustrade    is    divided    by    a 
series  of  vertical   iron  rods  into  equal 

rectangular  panels.  In  each  of  the  panels  an  iron  scroll 
like  the  one  shown  in  the  figure  is  constructed.  The 
rectangles  are  36  in.  high  and  12  in.  wide.  Construct  one 
of  the  panels,  using  a  convenient  scale,  from  the  following 
data.  The  center  of  the  first  semicircle  is  A,  the  radius 
AB  being  3  in.  0  is  the  mid-point  of  AB  and  the  center 
of  the  second  semicircle,  the  radius  OC  being  4i  in.  From 
A  as  a  center,  with  radius  A  C,  draw  the  next  arc,  meeting 
a  tangent  drawn  to  it  from  P,  the  center  of  the  panel. 
If  AP  =  12  in.,  prove  that  Zx  =  30°. 

39.  If  the  balustrade  in  the  previous  problem  contains  20  panels, 
what  is  the  combined  length  of  the  scrolls  ? 


INDEX 


(References  are  to  page  numbers.) 


Abbreviations,  table  of,  71 

Abscissa,  177 

Acute  angle,  18 

Acute  triangle,  30 

Addition,  proportion  by,  233 

Adjacent  angles,  16,  21 

Alternate-exterior  angles,  94 

Alternate-interior  angles,  94 

Alternation,  233 

Altitude,  of  parallelogram,  133  ;  of 
pole  star,  170  ;  of  trapezoid,  133 ; 
of  triangle,  89,  181 

Angle,  14  ;  acute,  18  ;  base,  of  tri- 
angle, 31  ;  bisection  of,  58,  87  ; 
bisector  of,  17  ;  central,  47,  286  ; 
construction  of,  48  ;  generated, 
16  ;  inscribed,  164  ;  magnitude 
of,  16  ;  measurement  of,  22,  50, 
164  seq. ;  oblique,  18  ;  obtuse, 
18 ;  of  depression,  65 ;  of  eleva- 
tion, 65;  reflex,  18;  right,  17; 
round,  16  ;  straight,  16  ;  vertex, 
of  triangle,  31 ;  vertex  of,  14 

Angle-sum,  102  seq. 

Angles,  adjacent,  16,  21 ;  alternate- 
exterior,  94  ;  alternate-interior, 
94 ;  classification  of,  16 ;  com- 
plementary, 19  ;  conjugate,  19  ; 
corresponding,  94;  equal,  17,  48, 
76,  etc.  ;  exterior,  of  polygon, 
33,  109  ;  exterior,  of  transversal, 
94  ;  exterior,  of  triangle,  29,  102 ; 


homologous,  61,  75,  237  ;  interior, 
of  polygon,  33,  108  ;  interior,  of 
transversal,  94  ;  interior,  of  tri- 
angle, 29,  102  ;  methods  of  nam- 
ing, 14 ;  supplementary,  19 ; 
vertical,  19,  20,  21 

Antecedent,  197 

Antiparallels,  249 

Apothem,  294 

Arc,  37  ;  bisection  of,  58,  87  ;  inter- 
cepted, 47,  154  ;  length  of,  302  ; 
major,  47  ;  measurement  of,  50, 
164, 302 ;  minor,  47;  subtended,  47 

Arches,  190,  256,  322-325 

Archimedes,  316,  317 

Arcs,  complementary,  47  ;  conju- 
gate, 47  ;  supplementary,  47 

Area,  of  circle,  305,  310,  313;  of 
kite,  199  ;  of  parallelogram,  201  ; 
of  plane  figures,  193  seq.;  of  rec- 
tangle, 194,  197,  199;  of  right 
triangle,  199  ;  of  segment  of  cir- 
cle, 306 ;  of  square,  194 ;  of 
trapezoid,  206  ;  of  triangle,  203 

Arrangement  of  demonstration,  70 

Assumptions,  preliminary,  72 

Axial  symmetry,  53 

Axioms,  73  ;  of  existence  of  limit, 
312  ;  of  inequality,  124  ;  of  par- 
allels, 97  ;  of  superposition,  74 

Axis,  of  coordinates,  177  ;  of  sym- 
metry, 54,  56 


327 


328 


INDEX 


Base,  corresponding,  89 ;  of  isosceles 

triangle,  31 ;  of  trapezoid,  122 
Base  angles  of  isosceles  triangle,  31, 

79 
Bisection,  of  angle,  58,  87;  of  arc, 

58,  87 ;  of  line-segment,  11,  57,  86 
Bisector,  of  angle,  17  ;  of  angle  of 

triangle,  87,  137,  181,  280 
Brescia  arch,  324 

Center,  of  circle,  36;  of  gravity, 
139 ;  of  regular  polygon,  294 ;  of 
similitude,  250 

Center  line,  40 

Center  segment,  40 

Central  angle,  47,  48 

Centroid,  138 

Chord,  39,  145  ;  of  contact,  155 

Circle,  36,  143 ;  and  angle,  47  ;  arc 
of,  37  ;  area  of,  305 ;  center  of, 
36 ;  circumference  of,  36,  298, 
299,  310,  313;  circumscribed, 
145,  146;  diameter  of,  36; 
escribed,  157 ;  exterior  region  of, 
37  ;  inscribed,  155,  156  ;  interior 
region  of,  37 ;  mensuration  of, 
309 ;  radius  of,  30 ;  sector  of, 
306  ;  segment  of,  306 

Circles,  and  proportional  lines, 
265  ;  concentric,  40  ;  equal,  36  ; 
lines  and,  38 ;  points  and,  38  ; 
tangent,  159 ;  two,  39,  159 

Circumcenter,  136 

Circumference  of  circle,  36,  298, 
299,  310,  313 

Circumscribed  circle,  145,  146 

Circumscribed  polygon,  155,  285 

Circumscribed  triangle,  32,  156 

Collinearity,  134,  136 

Compass,  mariner's,  25 

Compasses,  37  ;  proportional,  253 

Complement,  19,  21,  47 


Composite  figures,  116,   141,  289, 

291,  319 

Concentric  circles,  40 
Conclusion,  70 
Concurrence,  134,  136 
Cone,  4,  253 
Congruence,  61,  74 ;  laws  of,  63 ;  of 

polygons,  83 ;  of  right  triangles,  91 
Congruent  figures,  61 
Conjugate  angles,  19 
Conjugate  arcs,  47 
Consequent,  197 
Constant,  309 
Construction,    of    angles,    48 ;    of 

axis,  56 ;  of  right  triangles,  91  ; 

of  symmetric  figures,  54  ;  of  tri- 
angles, 43,  44,  49,  181 
Constructions,    57,    84,    181    seq.; 

fundamental,  84 
Contact,  point  of,  39,  153,  159 
Continued  proportion,  231 
Converse  propositions,  80 
Convex  polygon,  33 
Coordinates,  177 
Corollary,  79 
Corresponding  angles,  94 
Cosine,  259 
Cross-section  paper,  177,  195,  207, 

219,  237,  251,  261,  305 
Cube,  4,  200,  218 
Curved  line,  3 
Cylinder,  4,  303,  308 

Decagon,  33,  290 

Degree,  of  angle,  22  ;  of  arc,  49 

Demonstration,  arrangement  of, 
70  ;  form  of,  70 

Demonstrative  geometry,  69 

Depression,  angle  of,  65 

Determination,  of  line,  8  ;  of  tri- 
angles, 183 

Determining  parts,  181 


INDEX 


329 


Diagonal  of  polygon,  33 

Diagonal  scale,  252 

Diameter,  36 

Dimensions,  194 

Directions,  opposite,  9,  11 

Distance,  between  two  parallel 
lines,  133  ;  between  two  points, 
9,  132  ;  from  a  point  to  a  line, 
132  ;  from  a  point  to  a  circle,  152 

Dodecagon,  33,  288 

Early  English  arch,  325 

Eclipse,  191 

Elevation,  angle  of,  65 

Equal  angles,  17,  48 

Equal  areas,  193 

Equal  circles,  36 

Equal  segments,  9,  11,  12 

Equation,  graph  of,  179 

Equiangular  triangle,  30,  80 

Equidistant  point,  43 

Equilateral  Gothic  arch,  190 

Equilateral  triangle,  44,  79 

Escribed  circle,  157 

Euclid,  2,  44       • 

Excenter,  137 

Exterior  angles,  of  polygon,  33, 109 ; 

of  transversal,  94  ;    of  triangle, 

29,  102 

Exterior  region  of  circle,  37 
Extreme  and  mean  ratio,  267,  268 
Extremes  of  a  proportion,  231 

Foot  of  perpendicular,  17 
Fourth  proportional,  231,  241 
Fundamental  constructions,  84 
Fundamental  operations,  11 


69; 


Gauss,  291 

Geometry,      demonstrative, 

Egyptian,  1  ;  Greek,  2  ;  method 
of,  5 ;  origin  of,  1  ;  purpose  of, 
3  ;  value  of,  4 


Golden  section,  269 
Gothic  arch,  190 
Graph  of  equation,  179 

Hero's  formula,  225 

Hexagon,  33,  288 

Hippocrates,  317 

Homologous  angles  and  sides,  61, 

75,  237 

Horseshoe  arch,  323 
Hypotenuse,  30 
Hypothesis,  70 

Incenter,  137 
Indirect  method,  95 
Inequalities,  124 
Inscribed  angle,  164 
Inscribed  circle,  155,  156 
Inscribed  polygon,  145,  285 
Inscribed  square,  250,  287 
Inscribed  triangle,  32,  51,  146 
Intercepted  arc,  47,  154 
Intersection  of  loci,  173 
Isosceles  trapezoid,  122 
Isosceles  triangle,  31,  43,  56,  79 
Interior  angles,  of  polygon,  33, 108  ; 

of  transversal,  94 ;    of  triangle, 

29,  102 

Interior  region  of  circle,  37 
Inversely  proportional,  231 
Inversion,  proportion  by,  233 

Kite,  57,  85 ;  area  of,  199 
Kochansky,  317 

Latitude,  170,  264 

Legs,  of  isosceles  triangle,  31  ;   of 

right  triangle,  30 ;  of  trapezoid, 

122 

Lever,  234 

Limit  of  variable,  310 
Line,  3  ;    center,    40  ;    curved,   3  ; 

straight,  3,  6,  7 


330 


INDEX 


Line-segment,  7,  9,  11 

Lines,  and  circles,  38;  concurrent, 

136;  parallel,  94  scq.,  119,  158; 

perpendicular,  17,  57,  58,  88,  89, 

90 

Locus  (loci)  ,171;  intersection  of,  1 73 
Locus  problems,  171-176,  192,  284  ; 

rules  for  solving,  171 

Mariner's  compass,  25 
Mean  proportional,  231,  255,  266 
Means  of  a  proportion,  231 
Measurement,  of  angles,  22, 50, 164 ; 

of  areas,  194  ;  of  line-segments,  9 
Median  of  triangle,  86,  181,  279 
Mensuration  of  circle,  309 
Mid-line  of  trapezoid,  122 
Mid-point,  11 
Minutes  of  angle,  22 
Mirror,  reflection  in,  59,  107,  129, 

133,  142,  147,  157 
Moldings,  189,  325 
Multifoils,  321 

Normal,  157 

Oblique  angles,  18 

Oblique  lines,  18 

Oblique  triangles,  30 

Obtuse  angle,  18 

Obtuse  triangle,  30 

Octagon,  33 

Opposite  directions,  11 

Opposite  sides,  11 

Optical  illusions,  9,  101 

Ordinate,  177 

Origin,  of  coordinates,  177;  of  ray,  11 

Orthocenter,  139 

Pantograph,  252 
Pappus,  theorem  of,  213 
Parallel   lines,   94  seq.,  119,  158; 
axiom  of,  97  ;  construction  of,  97 


Parallel  of  latitude,  264 

Parallel  ruler,  117 

Parallelogram,  111  seq.;  altitude 
of,  133 ;  area  of,  201 

Parallelogram  of  velocities,  117 

Parquet  flooring,  35 

Parts,  determining,  181  ;  principal, 
61, 181 ;  relation  of,  182 ;  second- 
ary, 181 

Pedal  triangle,  249 

Pentadecagon,  290 

Pentagon,  33,  290 

Penumbra,  191 

Perigon,  16 

Perimeter,  28,  33 

Perpendicular,  17,  90;  construc- 
tion of,  57,  58,  88,  89 

Perpendicular  bisector,  57,  85 

Persian  arch,  324 

Pi  (TT),  302  ;  history  of,  316 

Plane,  3,  6 

Plato,  84 

Plotting  of  points,  177 

Point,  3  ;  locus  of,  171  ;  of  contact 
or  tangency,  39,  153,  159;  plot- 
ting of ,  177 

Points,  and  circles,  38 ;  symmetric, 
54 

Polygon,  33;  angles  of,  33,  108, 
109;  circumscribed,  155,  285; 
convex,  83 ;  diagonals  of,  33  ; 
inscribed,  145,  285 ;  interior  an- 
gles of,  33,  108  ;  regular,  34,  50, 
88,  108,  285  seq. ;  sides  of,  33 

Polygons,  33  ;  congruence  of,  83 ; 
mutually  equiangular,  237  ;  simi- 
lar, 236  seq.,  271  seq. 

Preliminary  propositions,  72,  143, 
199,  232,  286 

Principal  elements  or  parts  of  tri- 
angle, 61,  181 

Prism,  4,  223,  224,  230 


INDEX 


331 


Projection  of  line,  276 

Proof,  form  of,  70 

Proportion,  197,  231  seq. 

Proportional,  fourth,  231,  241  ; 
mean,  231,  255,  266 ;  third,  231 
'roportional,  inversely,  231 

Proportional  compasses,  253 

Proportional  segments,  240,  241, 
242 

Propositions,  a  system  of,  69  ;  pre- 
liminary, 72,  143 

Protractor,  22,  49 

Ptolemy,  292 

Pyramid,  4,  204,  223 

Pythagoras,  theorem  of,  214  seg., 
255 


Quadrant,  37 

Quadrilateral,  33  ;  construction  of, 

186 
Quatrefoil,  320 

Radius,  of  circle,  36  ;  of  circum- 
scribed circle,  181,  281;  of  in- 
scribed circle,  181,  282 ;  of  regular 
polygon,  294 

Ratio,  197 ;  extreme  and  mean, 
267,  268;  of  similitude,  238; 
trigonometric,  259 

Ray,  11 

Rectangle,  113;  area  of,  194,  197, 
199 

Rectilinear  figures,  69 

Reflection  in  mirror,  59,  107,  129, 
133,  142,  147,  157 

Reflex  angle,  18 

Regular  polygon,  34,  50,  88,  108, 
285  seq. 

Regular  tetrahedron,  223 

Rhomboid,  113 

Rhombus,  113  ;  area  of,  199 

Right  angle,  17 


Right  triangle,  30,  91  sec/.,  214; 
angles  of,  103 ;  area  of,  199 ; 
congruence  of,  91  seq. ;  construc- 
tion of,  91  ;  in  drawing,  103 

Round  angle,  16 

Scale,  8  ;  diagonal,  252 

Scalene  triangle,  43 

Scheme,  76,  82,  93,  118,  248 

Scrolls,  325 

Secant,  39,  145,  167,  265,  266 

Secondary  parts  of  triangle,  181 

Seconds  of  angle,  22 

Sector  of  circle,  306 

Segment,  center,  40  ;  line-,  7,  9,  11 ; 
of  circle,  306 

Segmental  pointed  arch,  324 

Segments,  equal,  9, 11,  12  ;  propor- 
tional, 240  seq. 

Semicircle,  37,  147 

Sextant,  142 

Shadow,  190,  191,  236 

Side,  of  angle,  14  ;  of  polygon,  33  ; 
of  triangle,  28 

Sides,  homologous,  61,  75,  237 

Similar  polygons,  236  seq.,  271  seq. 

Similar  triangles,  244  seq. 

Similitude,  center  of,  250 ;  ratio  of, 
238 

Simpson's  rule,  208 

Sine,  259 

Snow  crystals,  55 

Solids,  geometric,  3,  4,  37,  45,  200, 
202,  204,  206,  218,  219,  223,  224, 
228,  230,  253,  303,  304,  308,  322 

Space,  3 

Sphere,  4 

Square,  113 ;  area  of,  194;  inscribed, 
250,  287 

Squared  paper,  177,  195,  207,  219, 
237,  251,  261,  305 

Standards,  8,  17 


332 


INDEX 


Steam  engine,  mechanism  of,  128 
Straight  angle,  16 
Straight  line,  3,  6,  7 
Subtended  arc,  47 
Subtraction,  proportion  by,  233 
Superposition,      axiom      of,      74 ; 

method  of,  62,  74 
Supplement,  19,  21,  47 
Surfaces,  plane  and  curved,  3 
Surveying,  65, 236, 251,  256, 262  seq. 
Symbols,  table  of,  71 
Symmetric,  54 
Symmetric  points,  54 
Symmetry,  axial,  53  ;  axis  of,  54,  56 

T-square,  100 

Tangency,  point  of,  39,  153,  159 

Tangent,  39,  153  seq.,  266;  com- 
mon, 161  ;  construction  of,  154, 
161 ;  trigonometric,  259 

Tangent  circles,  159 

Tetrahedron,  regular,  223 

Thales  of  Miletus,  68,  251 

Theorem,  of  Pappus,  213 ;  of 
Pythagoras,  214  sec;.,  255  ;  rules 
for  demonstration  of,  70 

Third  proportional,  231 

Three-point  problem,  176 

Transformations,  of  plane  figures, 
193,  210  seq. ;  of  proportions,  233 

Transit,  65 

Transversal,  94 

Transversal  theorem,  119 

Trapezium,  122 

Trapezoid,  122  ;  altitude  of,  133  ; 
area  of,  206  ;  isosceles,  122 


Trapezoidal  rule,  207 

Trefoils,  319 

Triangle,  28;  acute,  30;  altitude 
of,  89,  181;  angles  of ,  28  ;  area 
of,  203  ;  circumscribed,  32,  156  ; 
construction  of,  43,  44,  49,  181  ; 
determination  of,  183  ;  equiangu- 
lar, 30,  80  ;  equilateral,  44,  79 ; 
exterior  angles  of,  29,  102  ;  in- 
scribed, 32,  51,  146;  interior 
angles  of,  29,  102 ;  isosceles,  31, 
43,  56,  79;  median  of,  86,  181, 
279 ;  numerical  properties  of, 
276  seq.;  oblique,  30;  obtuse,  30; 
pedal,  249 ;  right,  30,  91  seq.,  214 ; 
scalene,  43 ;  sides  of,  28 ;  vertices 
of,  28 

Triangles,  classification  of,  30 ; 
similar,  244  seq. 

Trigonometric  ratios,  259  seq. 

Trigonometric  table,  260 

Umbra,  191 

Units,  of  angular  magnitude,  22  ; 
of  area,  193  ;  of  length,  8 

Variable,  309 

Velocities,  parallelogram  of,  117 

Vertex,  of  angle,  14  ;   of  triangle, 

28 

Vertex  angle  of  triangle,  31 
Vertical  angles,  19,  20,  21 

Wheel  and  axle,  303 
Whispering  gallery,  157 


ANNOUNCEMENTS 


FIRST  COURSE   IN   ALGEBRA 

By  HERBERT  E.  HAWKES,  Professor  of  Mathematics  in  Columbia  Uni- 
versity, WILLIAM  A.  LUBY  and  FRANK  C.  TOUTON,  Instructors 
in  Mathematics,  Central  High  School,  Kansas  City,  Mo. 


I2.mo,  cloth,  vii  +334  pages,  illustrated,  list  price,  $1.00 


QELDOM  has  a  textbook  met  with  such  signal  success  at  its  first  ap- 
^  pearance  as  Hawkes,  Luby,  and  Teuton's  "  First  Course  in  Algebra." 
Following  are  some  of  the  features  on  which  this  success  is  founded : 

1.  Sanity 

It  embodies  a  one  year's  course  which  is  thorough,  but  not  too  difficult,  with 
due  attention  to  the  really  valuable  recent  developments  in  the  teaching  of 
algebra.  No  hobby  is  ridden. 

2.  Careful  Gradation 

The  topics,  drill  exercises,  and  problems  were  selected  and  arranged  with 
the  greatest  care  and  with  a  constant  regard  for  the  ability  of  average  pupils. 

2.    Balance  of  Technic  and  Reasoning  Power 

Ample  drill  in  the  elementary  technic  of  algebra  is  accompanied  by  a  com- 
mensurate development  of  reasoning  power. 

4.  Wealth  of  Illustrative  Material 

In  the  explanation  of  points  which  experience  has  shown  to  be  difficult  for 
students  to  grasp,  there  is  an  abundance  of  illustrative  material  which  assists 
the  pupil  to  understand  the  subject. 

5.  Abundance  of  Carefully  Selected  Problems 

Great  care  is  given  the  important  question  of  problems.  The  large  num- 
ber included  in  the  book  will  be  found  practical,  interesting,  —  as  they  bear 
on  the  pupil's  everyday  life,  —  and  unusually  well  graded. 

6.  Reference  to  Arithmetic 

In  the  explanation  of  algebraic  processes  the  student's  confidence  is  estab- 
lished by  constant  references  to  the  already  familiar  subject  of  arithmetic. 

7.  Correlation  with  Geometry  and  Physics 

The  choice  of  topics  and  their  treatment  was  determined  by  the  fact  that 
many  students  now  study  geometry  and  physics  after  one  year's  work  in 
algebra. 

8.  Sensible  Treatment  of  Factoring 

Only  the  simpler  types  of  factors  are  considered.  Many  examples  give  the 
student  timely  assistance  with  the  numerous  difficulties  which  necessarily 
arise.  The  frequent  lists  of  review  exercises  in  factoring  should  give  a 
secure  grasp  of  forms  and  methods  in  the  shortest  possible  time. 


GINN  AND  COMPANY   PUBLISHERS 


SECOND   COURSE    IN   ALGEBRA 

By  HERBERT  E.  HAWKES,  Professor  of  Mathematics  in  Columbia  University, 

WILLIAM  A.  LUBY,  Head  of  the  Department  of  Mathematics,  Central 

High  School,  Kansas,  Mo.,  and  FRANK  C.  TOUTON,  Principal 

of  Central  High  School,  St.  Joseph,  Mo. 

I2mo,  cloth,  viii  +  264  pages,  illustrated,  75  cents 

THIS  book  is  designed  to  follow  the  authors'  "  First  Course 
in  Algebra"  and  continues  the  distinctive  methods  — 
liberal  use  of  illustrative  material,  introduction  of  numerous 
interesting  and  "  thinkable "  problems,  correlation  with  arith- 
metic, geometry,  and  physics,  and  extended  work  with  graphs 
—  which  marked  the  earlier  volume.  As  in  the  "  First  Course," 
prominence  is  given  the  equation ;  the  habit  of  checking  results 
is  constantly  encouraged ;  and  frequent  short  reviews  are  a 
feature  throughout. 

The  earlier  chapters  present  a  brief  but  thorough  review  of 
the  first-year  work,  giving  each  topic  a  broader  and  more 
advanced  treatment  than  is  permissible  in  the  "  First  Course." 
The  new  material  and  the  many  new  applications  make  the 
entire  review  appeal  to  the  student  as  fresh  and  inviting.  The 
later  chapters  introduce  such  further  topics  as  progressions, 
limits  and  infinity,  ratio  and  proportion,  logarithms,  and  the 
binomial  theorem. 

The  aim  throughout  has  been  to  select  those  topics  consid- 
ered necessary  for  the  best  secondary  schools  and  to  treat  each 
in  a  clear,  practical,  and  attractive  manner.  The  authors  have 
sought  to  prepare  a  text  that  will  lead  the  student  to  think 
clearly  as  well  as  to  acquire  the  necessary  facility  on  the 
technical  side  of  algebra. 

119  b 

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SHOP  PROBLEMS  IN   MATHE 
MATICS 

By  WILLIAM  E.  BRECKENRIDGE,  Chairman  of  the  Department  of  Mathematics, 

SAMUEL  F.  MERSEREAU,  Chairman  of  the  Department  of  Woodworking, 

and  CHARLES  F.  MOORE,  Chairman  of  the  Department  of  Metal 

Working  in  Stuyvesant  High  School,  New  York  City 

A?iswer  Book  furnished  on  application 


Cloth,  I2mo,  280  pages,  illustrated,  $1.00 


THIS  book  aims  to  give  a  thorough  training  in  the  mathe- 
matical operations  that  are  useful  in  shop  practice,  e.g.  in 
Carpentry,  Pattern  Making  and  Foundry  Work,  Forging,  and 
Machine  Work,  and  at  the  same  time  to  impart  to  the  student 
much  information  in  regard  to  shops  and  shop  materials.  The 
mathematical  scope  varies  from  addition  of  fractions  to  natural 
trigonometric  functions.  The  problems  are  practical  applica- 
tions of  the  processes  of  mathematics  to  the  regular  work  of 
the  shop.  They  are  graded  from  simple  work  in  board  measure 
to  the  more  difficult  exercises  of  the  machine  shop.  Through 
them  students  may  obtain  a  double  drill  which  will  strengthen 
their  mathematical  ability  and  facilitate  their  shop  work. 

All  problems  are  based  on  actual  experience.  The  slide  rule  is 
treated  at  length.  Short  methods  and  checks  are  emphasized. 
Clear  explanations  of  the  mechanical  terms  common  to  shop 
work  and  illustrations  of  the  machinery  and  tools  referred  to  in 
the  text  make  the  book  an  easy  one  for  both  student  and 
teacher  to  handle. 

It  should  be  useful  in  any  school,  elementary  or  advanced, 
where  there  are  shops,  as  a  review  for  supplementary  work  or 
as  a  textbook  either  in  mathematics  or  shop  work. 

132^ 

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TEXTBOOKS   IN   MATHEMATICS 

FOR    HIGH    SCHOOLS    AND    ACADEMIES 

Baker,  A. :  Elementary  Plane  Geometry $0.50 

Baker,  A.  L. :  Elements  of  Solid  Geometry 80 

Beman  and  Smith  :  Academic  Algebra 1.12 

Beman  and  Smith  :  Elements  of  Algebra 1.12 

Beman  and  Smith  :  Higher  Arithmetic 80 

Beman  and  Smith:  New  Plane  and  Solid  Geometry  .     .      1.25 
Breckenridge,  Mersereau,  and  Moore:  Shop  Problems 

in  Mathematics i.oo 

Cobb  :   Elements  of  Applied  Mathematics i.oo 

Hawkes  :  Advanced  Algebra 1.40 

Hawkes,  Luby,  and  Touton  :  First  Course  in  Algebra    .      i.oo 
Hawkes,  Luby,  and  Touton:   Second  Course  in  Algebra       .75 

Robbins :  Algebra  Reviews       . 25 

Smith:  Algebra  for  Beginners 50 

Smith:  The  Teaching  of  Geometry 1.25 

Wentworth :  Advanced  Arithmetic i.oo 

Wentworth  :   Elementary  Algebra 1.12 

Wentworth  :   First  Steps  in  Algebra 60 

Wentworth:  Higher  Algebra 1.40 

Wentworth  :  New  School  Algebra 1.12 

Wentworth  :  Plane  and  Solid  Geometry  (Revised,  WTent- 

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Plane  Geometry  (Revised,  Wentworth-Smith  Series)    .       .80 

Solid  Geometry  (Revised, Wentworth-Smith  Series)     .       .75 

Wentworth  and  Hill:  Exercise  Manual  in  Geometry       .       .70 

Wentworth  and  Hill :  Exercises  in  Algebra 70 

Wentworth  and  Hill :   First  Steps  in  Geometry      ...       .60 


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